Question

$2\,cos^{-1}\,x = sin^{-1}\left(2x\sqrt{1-x^{2}}\right)$ is true for

• A. all $x$
• B. $x > 0$
• C. $x \in [-1,1]$
• D. $\frac{1}{\sqrt{2}} \le x \le 1$

Answer 1

Answer: (D)

$2\,cos^{-1}\,x = sin^{-1}2x\sqrt{1-x^{2}} $
$\therefore $ Range of $sin^{-1}x$ function is $\left[-\pi/2,\pi/2\right]$
$\Rightarrow -\frac{\pi}{2} \le sin^{-1}\left(2x\sqrt{1-x^{2}}\right)\le \frac{\pi}{2}$
$\Rightarrow -\frac{\pi }{2} \le 2\,cos^{-1}\,x \le \frac{\pi }{2}$
$\Rightarrow -\frac{\pi }{4} \le cos ^{-1}\,x \le \frac{\pi }{4}\quad\ldots\left(i\right)$
$\therefore $ Range of $cos^{-1}x$ is $\left[0, \pi\right]$.
Hence from $\left(i\right)$
$0 \le cos^{-1}x \le \frac{\pi}{4}$
$\Rightarrow 1 \ge x \ge \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{1}{\sqrt{2}} \le x \le 1$