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Question

Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]

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Solution

We know that 
\[\sin^{- 1} \left( \sin{x} \right) = x\]
We have
\[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right) = \sin^{- 1} \left\{ \sin\left( \pi - \frac{3\pi}{5} \right) \right\} \left[ \because \left( \pi - \frac{3\pi}{5} \right) \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ = \sin^{- 1} \left( \sin\frac{2\pi}{5} \right)\]
\[ = \frac{2\pi}{5}\]
∴ \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right) = \frac{2\pi}{5}\]

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Chapter 4: Inverse Trigonometric Functions - Exercise 4.15 [Page 118]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.15 | Q 32 | Page 118

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