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

Answers (1)

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}

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