sin^{-1}(sin frac{7pi}{6}) = | vedclass

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Question

(C) The principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $\frac{7\pi}{6}$ does not lie in the interval $[-\frac{\pi}

Vedclass · Accepted Answer

$-\frac{\pi}{6}$

Vedclass · Answer

(C) The principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $\frac{7\pi}{6}$ does not lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we simplify the expression.We know that $\sin(\pi + 	heta) = -\sin 	heta$,but it is easier to use $\sin(\pi + 	heta) = \sin(2\pi - (\pi + 	heta)) = \sin(\pi - 	heta)$ is not correct here.Instead,use $\sin(\frac{7\pi}{6}) = \sin(\pi + \frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$.Thus,$\sin^{-1}(\sin \frac{7\pi}{6}) = \sin^{-1}(\sin(-\frac{\pi}{6}))$.Since $-\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the value is $-\frac{\pi}{6}$.

$\sin^{-1}(\sin \frac{7\pi}{6}) = $

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