Answer
$\dfrac{\sqrt 3}{2}$
Work Step by Step
The range of $\cos^{-1}{x}$ is $[0, \pi]$.
Here, $\cos^{-1}{(\dfrac{1}{2})}=\dfrac{\pi}{3}$, because $\cos{(\dfrac{\pi}{3})}=\dfrac{1}{2}$ and $\dfrac{\pi}{3}$ lies in the range of $\cos^{-1}{x}$ .
Therefore, we have:
$\sin [\cos^{-1}{(\dfrac{1}{2})}]=\sin ({\dfrac{\pi}{3}})=\dfrac{\sqrt 3}{2}$