Question

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

• A. $x\in [0,1]$
• B. $x\in \left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$
• C. $x\in (-1,0)$
• D. $x\in \left[-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}\right]$

Answer 1

Answer: (B)

Let $x=sin\theta$
Then $si{n}^{-1}\left(2x\sqrt{1-x^2}\right)=si{n}^{-1}\left(sin2\theta \right)$
$=2\theta =2si{n}^{-1}x$
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 $-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}$
$\therefore x\in \left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$