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Difference between revisions of "Euc20198/Sub-Problem 2"

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== Problem ==
 
== Problem ==
  
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Given <math>0<x<\frac{\pi}{2}</math> and <math>\cos(\frac{3}{2}\cos(x))</math> = <math>\sin(\frac{3}{2}\sin(x))</math>, determine <math>\sin(2x)</math>, represented in the form (a(<math>\pi</math>)^2 + b(<math>\pi</math>) + c)/d where a, b, c, d are integers.
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Given <math>0<x<\frac{\pi}{2}</math> and <math>\cos(\frac{3}{2}\cos(x))</math> = <math>\sin(\frac{3}{2}\sin(x))</math>, determine <math>\sin(2x)</math>, represented in the form <math>\frac{a\pi^2 + b\pi + c}{d}</math> where a, b, c, d are integers.
  
 
== Solution ==
 
== Solution ==
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First, we use the identity that <math>\cos \theta=\sin(\frac{\pi}{2}-\theta)</math> on the left hand side of the equation, so the equation becomes <math>\sin(\frac{\pi}{2}-\frac{3}{2}\cos(x))=\sin(\frac{3}{2}\sin x)</math>. Both arguments to sine are in <math>(0,\pi)</math>, so we can equate them to <math>\frac{\pi}{2}-\frac{3}{2}\cos x=\frac{3}{2}\sin x</math>. Multiplying each side by 2, we get <math>\pi-3\cos x=3\sin x</math>. Rewriting the equation gives us <math>3\cos x+3\sin x=\pi</math> so dividing by 3 gives us <math>\cos x+\sin x=\frac{\pi}{3}</math>. Notice that, if we square both sides, we get <math>\cos^2 x+2\sin x\cos x+\sin^2 x=\frac{\pi^2}{9}</math>, and by using the identity that <math>\cos^2 x+\sin^2 x=1</math>, we get <math>1+2\sin x\cos x=\frac{\pi^2}{9}</math>, and notice that <math>\sin(2x)=\sin(x+x)=\sin x\cos x+\cos x\sin x=2\sin x\cos x</math>, which is in our equation, so <math>1+\sin(2x)=\frac{\pi^2}{9}</math> and subtracting 1 on both sides gives <math>\sin(2x)=\frac{\pi^2}{9}-1</math>. This means that <math>\sin(2x)=\frac{\pi^2-9}{9}</math>, so our final answer is <math>\boxed{\frac{\pi^2-9}{9}}</math>, where <math>a=1</math>, <math>b=0</math>, <math>c=-9</math>, and <math>d=9</math>.
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~Baihly2024
  
 
== Video Solution ==
 
== Video Solution ==

Latest revision as of 16:30, 12 October 2025

Problem

Given $0<x<\frac{\pi}{2}$ and $\cos(\frac{3}{2}\cos(x))$ = $\sin(\frac{3}{2}\sin(x))$, determine $\sin(2x)$, represented in the form $\frac{a\pi^2 + b\pi + c}{d}$ where a, b, c, d are integers.

Solution

First, we use the identity that $\cos \theta=\sin(\frac{\pi}{2}-\theta)$ on the left hand side of the equation, so the equation becomes $\sin(\frac{\pi}{2}-\frac{3}{2}\cos(x))=\sin(\frac{3}{2}\sin x)$. Both arguments to sine are in $(0,\pi)$, so we can equate them to $\frac{\pi}{2}-\frac{3}{2}\cos x=\frac{3}{2}\sin x$. Multiplying each side by 2, we get $\pi-3\cos x=3\sin x$. Rewriting the equation gives us $3\cos x+3\sin x=\pi$ so dividing by 3 gives us $\cos x+\sin x=\frac{\pi}{3}$. Notice that, if we square both sides, we get $\cos^2 x+2\sin x\cos x+\sin^2 x=\frac{\pi^2}{9}$, and by using the identity that $\cos^2 x+\sin^2 x=1$, we get $1+2\sin x\cos x=\frac{\pi^2}{9}$, and notice that $\sin(2x)=\sin(x+x)=\sin x\cos x+\cos x\sin x=2\sin x\cos x$, which is in our equation, so $1+\sin(2x)=\frac{\pi^2}{9}$ and subtracting 1 on both sides gives $\sin(2x)=\frac{\pi^2}{9}-1$. This means that $\sin(2x)=\frac{\pi^2-9}{9}$, so our final answer is $\boxed{\frac{\pi^2-9}{9}}$, where $a=1$, $b=0$, $c=-9$, and $d=9$.

~Baihly2024

Video Solution

https://www.youtube.com/watch?v=3ImnLWRcjYQ

~NAMCG

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