sin^{-1} (sin frac{3pi}{5}) = dots dots dots | vedclass

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Question

(A) The principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $\frac{3\pi}{5} > \frac{\pi}{2}$, we must rewrite $\sin(\fr

Vedclass · Accepted Answer

$\frac{2\pi}{5}$

Vedclass · Answer

(A) The principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $\frac{3\pi}{5} > \frac{\pi}{2}$, we must rewrite $\sin(\frac{3\pi}{5})$ using the identity $\sin(\pi - 	heta) = \sin(	heta)$.Thus, $\sin(\frac{3\pi}{5}) = \sin(\pi - \frac{3\pi}{5}) = \sin(\frac{2\pi}{5})$.Since $\frac{2\pi}{5}$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we have $\sin^{-1}(\sin \frac{2\pi}{5}) = \frac{2\pi}{5}$.

$\sin^{-1} (\sin \frac{3\pi}{5}) = \dots \dots \dots$

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