Question

Express ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ in terms of ${\sin }^{-1}x$

Answer 1

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

$\Rightarrow x=\sin \theta$ and $\frac{-\pi }{2}\le \theta \le \frac{\pi }{2}$

i.e., $-\pi \le 2\theta \le \pi$

Consider the following cases:

When $\frac{-\pi }{2}\le 2\theta \le \frac{\pi }{2}$

i.e.,$\frac{-1}{\sqrt{2}}\le x\le \frac{1}{2}$

the given expression may be written as

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

$=2\theta$

$=2{\sin }^{-1}x$

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