Answer
$\dfrac{\sqrt2}{2}$
Work Step by Step
The range of $\sin^{-1}{x}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.
Here, $\sin^{-1}{\left(\dfrac{\sqrt2}{2}\right)}=\dfrac{\pi}{4}$, because $\sin{\left(\dfrac{\pi}{4}\right)}=\dfrac{\sqrt2}{2}$ and $\dfrac{\pi}{4}$ lies in the range of $\sin^{-1}{x}$ .
Therefore, we have:
$\cos{\left(\sin^{-1}{\left(\dfrac{\sqrt2}{2}\right)}\right)}=\cos ({\dfrac{\pi}{4}})=\dfrac{\sqrt2}{2}$
