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  • Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 32 math

Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 32 math

Answers (1)

Answer: \frac{2 \pi}{5}

Given:

\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)

Hint:

\sin ^{-1}(\sin x)=x

Solution:             

\begin{array}{l} \sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left\{\sin \left(\pi-\frac{3 \pi}{5}\right)\right\} \quad:\left[\left(\pi-\frac{3 \pi}{5}\right) \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\right] \\\\ \end{array}

                                 \qquad \begin{array}{l} =\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right) \\\\ =\frac{2 \pi}{5} \end{array}

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