If frac{pi}{2} le x le frac{3pi}{2}, then sin | vedclass

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(D) We know that the range of the principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Given $\frac{\pi}{2} \le x \le \frac{3\

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$\pi - x$

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(D) We know that the range of the principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Given $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$.We use the property $\sin x = \sin(\pi - x)$.Since $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$,we have $-\frac{\pi}{2} \le \pi - x \le \frac{\pi}{2}$.Therefore,$\sin^{-1}(\sin x) = \sin^{-1}(\sin(\pi - x)) = \pi - x$.

If $\frac{\pi}{2} \le x \le \frac{3\pi}{2},$ then $\sin^{-1}(\sin x)$ is equal to

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