Answer
$\dfrac{1}{2}$
Work Step by Step
The range of $\sin^{-1}{x}$ is $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$.
Here, $\sin^{-1}{(\dfrac{-\sqrt 3}{2})}=\dfrac{-\pi}{3}$, because $\sin{(\dfrac{-\pi}{3})}=\dfrac{-\sqrt 3}{2}$ and $\dfrac{-\pi}{3}$ lies in the range of $\sin^{-1}{x}$ .
Therefore, we have:
$\cos [\sin^{-1}{(\dfrac{-\sqrt 3}{2})}]=\cos ({\dfrac{-\pi}{3}})=\dfrac{1}{2}$
