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Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 62

Answers (1)

Answer:  \frac{-\pi}{8}

Given:

\sin ^{-1}\left[\left(\sin -\frac{17 \pi}{8}\right)\right]

Hint:

\sin \left(\sin ^{-1} \theta\right)=\theta \ \ if \ \ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}

Solution:        

  \begin{aligned} \sin ^{-1}\left(\sin -\frac{17 \pi}{8}\right) \sin ^{-1}\left(-\sin \frac{17 \pi}{8}\right) &=\sin ^{-1}\left[-\sin \left(2 \pi+\frac{\pi}{8}\right)\right] \\ &=\sin ^{-1}\left(-\sin \frac{\pi}{8}\right) \\ &=\sin ^{-1}\left(\sin -\frac{\pi}{8}\right) \\ &=-\frac{\pi}{8} \end{aligned}

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