Question

Find the values of $x$ for which ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=2{\sin }^{-1}x$ holds.

Answer 1

Let $x=\sin \theta \Rightarrow \theta ={\sin }^{-1}x$

Therefore,

${\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)=2{\sin }^{-1}\left(\sin \theta \right)$

${\sin }^{-1}\left(2\sin \theta \sqrt{{\cos }^2\theta }\right)=2\theta (\because {\sin }^{-1}\left(\sin \alpha \right)=\alpha )$

${\sin }^{-1}\left(2\sin \theta \cos \theta \right)=2\theta$

${\sin }^{-1}\left(\sin 2\theta \right)=2\theta$

$2\theta =2\theta$

L.H.S. $=$ R.H.S.

Hence the given expression holds for $x=\sin \theta$ means $-1\le x\le 1$