Question

$2{\cos }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ is valid for all values of $x$ satisfying

• A. $-1\le x\le 1$
• B. $0\le x\le 1$
• C. $\frac{1}{\sqrt{2}}\le x\le 1$
• D. $0\le x\le \frac{1}{\sqrt{2}}$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}\le x\le 1$

Given equation is an identity, except the range should be same for both sides

The range of ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$

And the range of $2{\cos }^{-1}x$ is $[0,2\pi ]$

So the common range is $[0,\frac{\pi }{2}]$

$\Rightarrow 0\le 2{\cos }^{-1}x\le \frac{\pi }{2}$

$\Rightarrow 0\le {\cos }^{-1}x\le \frac{\pi }{4}$

$\Rightarrow \frac{1}{\sqrt{2}}\le x\le 1$, since ${\cos }^{-1}$ is decreasing function so, inequality will get reversed