Question

The formula
$2\,\sin^{-1}x=\sin^{-1}(2x\sqrt{1-x^2})$ holds for

• A. $x\,\in\,[0,1]$
• B. $x\,\in\,\left[-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right]$
• C. $x\,\in\,(-1,0)$
• D. $x\,\in\,\left[-\frac{\sqrt3}{2},\frac{\sqrt3}{2}\right]$

Answer 1

Answer: (B)

Let $x= sin\,\theta $
Then $sin^{-1}\left(2x\sqrt{1-x^{2}} \right) = sin^{-1} \left(sin \,2\theta\right) $
$ = 2\theta = 2\,sin^{-1} x$
if $-\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2}$ i.e., if $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4} $
i.e., if $sin\left(-\frac{\pi}{4}\right)\le sin \,\theta\le sin \frac{\pi}{4}$
i.e., if $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$
$\therefore x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$