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

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

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

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

(D) 1

Option D is correct

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

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