Question

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

Answer 1

Let $\cos \theta =x\Rightarrow \theta ={\cos }^{-1}x$
${\cos }^{-1}\left(2x\sqrt{1-x^2}\right)={\cos }^{-1}\left(2\cos \theta \sin \theta \right)$
$={\cos }^{-1}\left(\sin 2\theta \right)={\cos }^{-1}\left[\cos (\frac{\pi }{2}-2\theta )\right]=\frac{\pi }{2}-2\theta \Rightarrow {\cos }^{-1}\left(2x\sqrt{1-x^2}\right)=\frac{\pi }{2}-2{\cos }^{-1}x\text{(Proved)}$