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Question

If $0<x<\dfrac{1}{\sqrt2}$ and $\sin^{-1}\left(\dfrac{x}{\alpha}\right)=\cos^{-1}\left(\dfrac{x}{\beta}\right)$, then the value of $\sin\left(\dfrac{2\pi\alpha}{\alpha+\beta}\right)$ is

Accepted Answer

Let $\sin^{-1}\dfrac{x}{\alpha}=\cos^{-1}\dfrac{x}{\beta}=k$. Adding, $\sin^{-1}x+\cos^{-1}x=k(\alpha+\beta)\Rightarrow\alpha+\beta=\dfrac{\pi}{2k}$.

$$\frac{2\pi\alpha}{\alpha+\beta}=2\pi\alpha\cdot\frac{2k}{\pi}=4k\alpha=4\sin^{-1}x\quad(\text{since }k\alpha=\sin^{-1}x).$$

So we need $\sin(4\sin^{-1}x)$. Let $\theta=\sin^{-1}x$, $x=\sin\theta$, $\theta\in\left(0,\dfrac{\pi}{4}\right)$:

$$\cos2\theta=1-2x^2>0,\qquad \sin2\theta=2x\sqrt{1-x^2}.$$

$$\sin4\theta=2\sin2\theta\cos2\theta=2\cdot2x\sqrt{1-x^2}\cdot(1-2x^2)=4x\sqrt{1-x^2}\,(1-2x^2).$$

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