provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (vi)

Answer:   $-\frac{\pi }{8}$

Hint: The principal value branch of function  $\sin ^{-1}$  is  $\left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]$

Given:    $\sin ^{-1}\left\{\left(\sin -\frac{17 \pi}{8}\right)\right\}$

Explanation:

First we solve  $\left(\sin -\frac{17 \pi}{8}\right)$

As we know, $-\sin \theta=\sin (-\theta)$

\begin{aligned} &\therefore\left(\sin -\frac{17 \pi}{8}\right)=-\sin \frac{17 \pi}{8} \\ &-\sin \frac{17 \pi}{8}=-\sin \left(2 \pi+\frac{\pi}{8}\right) \end{aligned}

$-\sin \left(2 \pi+\frac{\pi}{8}\right)=-\sin \left(\frac{\pi}{8}\right) \quad \because[\sin (2 \pi+\theta)=\sin (\theta)]$

\begin{aligned} &\therefore[-\sin (\theta)=\sin (-\theta)] \\ &-\sin \left(\frac{\pi}{8}\right)=\sin \left(-\frac{\pi}{8}\right) \end{aligned}

By substituting these values in $\sin ^{-1}\left\{\left(\sin -\frac{17 \pi}{8}\right)\right\}$  we get,

$\sin ^{-1}\left(\sin -\frac{\pi}{8}\right)$

$\begin{gathered} \text { As } \sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ \therefore \sin ^{-1}\left(\sin -\frac{\pi}{8}\right)=-\frac{\pi}{8} \end{gathered}$