Question

Find the value of $\sin [\frac{\pi }{3}-{\sin }^{-1}(\frac{-1}{2}\left)\right]$.

Answer 1

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

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

$=\sin (\frac{\pi }{3}+\frac{\pi }{6})$

$=\sin \frac{3\pi }{6}=\sin \frac{\pi }{2}=1$