Question

If $x \in\left(\frac{3 \pi}{2}, 2 \pi\right)$, then the value of the expression $\sin ^{-1}\left[\cos \left\{\cos ^{-1}(\cos x)\right\} \sin ^{-1}(\sin x)\right]$, is

• A. $-\frac{\pi}{2}$
• B. $\frac{\pi}{2}$
• C. $0$
• D. $\pi$

Answer 1

Answer: (B)

$x \in\left(\frac{3 \pi}{2}, 2 \pi\right)$
Then, $\cos ^{-1}(\cos x)=2 \pi-x$
and $\sin ^{-1}(\sin x)=x-2 \pi$
$\therefore \cos ^{-1}(\cos x)+\sin ^{-1}(\sin x)=0$
Therefore, $\sin ^{-1}\left[\cos \left\{\cos ^{-1}(\cos x)+\sin ^{-1}(\sin x)\right\}\right]$
$=\sin ^{-1}\{\cos (0)\}=\sin ^{-1}(1)=\frac{\pi}{2}$