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  • Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 38

Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 38

Answers (1)

Answer:\frac{-\pi}{3}

Given:

\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)

Hint: \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)  is the range of the principal value branch of inverse sine function.

Solution:

Let \ \ y=\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)

Then,         

                 \begin{array}{c} \sin y=-\frac{\sqrt{3}}{2}=\sin \left(-\frac{\pi}{3}\right) \\\\ \end{array}

                \begin{array}{c} y=-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\\\ \end{array}

\begin{array}{c} \therefore \sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3} \end{array}

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