Question

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

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

Answer 1

$\left(i\right)L.H.S={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$
Substituting $x=\sin \theta$
${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$
$={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)$
$={\sin }^{-1}\left(2\sin \theta \sqrt{{\cos }^2\theta }\right)$
$={\sin }^{-1}\left(2\sin \theta |\cos \theta |\right)$
Given $-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}$, so
$\Rightarrow \theta ={\sin }^{-1}x\in [-\frac{\pi }{4},\frac{\pi }{4}]\Rightarrow \cos \theta >0$
${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)={\sin }^{-1}\left(2\sin \theta |\cos \theta |\right)$
Now,
$={\sin }^{-1}\left(2\sin \theta \cos \theta \right)$
$={\sin }^{-1}\left(\sin 2\theta \right)$
$\because -\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}$
$\Rightarrow \theta ={\sin }^{-1}x\in [-\frac{\pi }{4},\frac{\pi }{4}]$
$\Rightarrow 2\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$
$\Rightarrow {\sin }^{-1}\left(\sin 2\theta \right)$
$=2\theta =2{\sin }^{-1}x$
Hence proved.

$\left(ii\right)L.H.S={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$
Substituting $x=\cos \theta$
$\Rightarrow {\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$
$={\sin }^{-1}\left(2\cos \theta \sqrt{1-{\cos }^2\theta }\right)$
$={\sin }^{-1}\left(2\cos \theta \sqrt{{\sin }^2\theta }\right)$
$={\sin }^{-1}\left(2\cos \theta |\sin \theta |\right)$
Given $\frac{1}{\sqrt{2}}\le x\le 1$, so
$\Rightarrow \theta ={\cos }^{-1}x\in [0,\frac{\pi }{4}]\Rightarrow \sin \theta >0$
$={\sin }^{-1}\left(2\cos \theta \sin \theta \right)$
$={\sin }^{-1}\left(\sin 2\theta \right)$
$\because \frac{1}{\sqrt{2}}\le x\le 1$
$\Rightarrow \theta ={\cos }^{-1}x\in [0,\frac{\pi }{4}]$
$\Rightarrow 2\theta \in [0,\frac{\pi }{2}]$
$\Rightarrow {\sin }^{-1}\left(\sin 2\theta \right)$
$=2\theta =2{\cos }^{-1}x$
Hence proved.