Question

$2co{s}^{-1}x=si{n}^{-1}\left(2x\sqrt{1-x^2}\right)$ 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)

$2co{s}^{-1}x=si{n}^{-1}2x\sqrt{1-x^2}$
$\therefore$ Range of $si{n}^{-1}x$ function is $\left[-\pi /2,\pi /2\right]$
$\Rightarrow -\frac{\pi }{2}\le si{n}^{-1}\left(2x\sqrt{1-x^2}\right)\le \frac{\pi }{2}$
$\Rightarrow -\frac{\pi }{2}\le 2co{s}^{-1}x\le \frac{\pi }{2}$
$\Rightarrow -\frac{\pi }{4}\le co{s}^{-1}x\le \frac{\pi }{4}\dots \left(i\right)$
$\therefore$ Range of $co{s}^{-1}x$ is $\left[0,\pi \right]$.
Hence from $\left(i\right)$
$0\le co{s}^{-1}x\le \frac{\pi }{4}$
$\Rightarrow 1\ge x\ge \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{1}{\sqrt{2}}\le x\le 1$