Find the values of
Question:

Find the values of $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to

(A) $\frac{1}{2}$

(B) $\frac{1}{3}$

(C) $\frac{1}{4}$

(D) 1

Solution:

Let $\sin ^{-1}\left(\frac{-1}{2}\right)=x$. Then, $\sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right)$

We know that the range of the principal value branch of $\sin ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.

$\therefore \sin ^{-1}\left(\frac{-1}{2}\right)=\frac{-\pi}{6}$

$\therefore \sin \left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)=\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin \left(\frac{3 \pi}{6}\right)=\sin \left(\frac{\pi}{2}\right)=1$

The correct answer is $D$.

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