The value of sin−1⁡cos⁡3π5 is

=sin−1⁡cos⁡3π5=sin−1⁡cos⁡π2+π10=sin−1⁡−sin⁡π10 ∵cos⁡π2+x=−sin⁡x=−sin−1⁡sin⁡π10 ∵sin−1⁡(−x)=−sin−1⁡x=−π10∵sin−1⁡(sin⁡x)=x,x∈−π2,π2

The value of sin−1⁡cos⁡3π5 is

=sin−1⁡cos⁡3π5=sin−1⁡cos⁡π2+π10=sin−1⁡−sin⁡π10 ∵cos⁡π2+x=−sin⁡x=−sin−1⁡sin⁡π10 ∵sin−1⁡(−x)=−sin−1⁡x=−π10∵sin−1⁡(sin⁡x)=x,x∈−π2,π2

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The value of $\sin^{- 1} (\cos \frac{3 π}{5})$ is

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$\frac{π}{10}$

$\frac{3 π}{5}$

$- \frac{π}{10}$

$\frac{- 3 π}{5}$

answer is C.

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Detailed Solution

$\begin{array}{l} = \sin^{- 1} [\cos (\frac{3 π}{5})] \\ = \sin^{- 1} [\cos (\frac{π}{2} + \frac{π}{10})] \\ = \sin^{- 1} (- \sin \frac{π}{10}) [∵ \cos (\frac{π}{2} + x) = - \sin x] \\ = - \sin^{- 1} (\sin \frac{π}{10}) [∵ \sin^{- 1} (- x) = - \sin^{- 1} x] \\ = - \frac{π}{10} [∵ \sin^{- 1} (\sin x) = x, x \in (\frac{- π}{2}, \frac{π}{2})] \end{array}$