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  • Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (v) maths textbook solution

Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (v) maths textbook solution

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:   

Explanation:

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

            \begin{aligned} &\sin \frac{17 \pi}{8}=\sin \left(2 \pi+\frac{\pi}{8}\right) \\ &\sin \left(2 \pi+\frac{\pi}{8}\right)=\sin \left(\frac{\pi}{8}\right) \quad \because[\sin (2 \pi+\theta)=\sin (\theta)] \end{aligned}                                                          

By substituting these values in \sin ^{-1}\left(\sin \frac{17 \pi}{8}\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{17 \pi}{8}\right)=\frac{\pi}{8} \end{gathered}

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