Euc20198/Sub-Problem 2

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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Euc20198/Sub-Problem 2

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