Answer
$\dfrac{1}{2}$
Work Step by Step
Note that $\sin^{-1}{\left(-\dfrac{\sqrt3}{2}\right)}=-\dfrac{\pi}{3}$, because $\sin{\left(-\dfrac{\pi}{3}\right)}=-\dfrac{\sqrt3}{2}$ and $-\dfrac{\pi}{3}$ is in the range of $\sin^{-1}{x}$, which is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$
Thus
$\cos{\left(\sin^{-1}{\left(-\dfrac{\sqrt3}{2}\right)}\right)}=\cos{\left(-\dfrac{\pi}{3}\right)}=\dfrac{1}{2}$
