Question

Prove that ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=2{\cos }^{-1}x,\frac{1}{\sqrt{2}}\le x\le 1$

Answer 1

$x=\cos \theta$

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

$-\frac{\pi }{2}\le {\sin }^{-1}\left(2x\sqrt{1-x^2}\right)\le \frac{\pi }{2}$

$-\frac{\pi }{2}\le 2{\cos }^{-1}x\le \frac{\pi }{2}$

$-\frac{\pi }{4}\le {\cos }^{-1}x\le \frac{\pi }{4}$

$x\in [\frac{1}{\sqrt{2}},1]$