Answer
Proof given below.
Work Step by Step
Let $\sin y=x$, and $y\in[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}]$.
Then, by definition of $\sin^{-1},\ y=\sin^{-1}x$.
Because of the way we defined y,
$y $ belongs to either the 1st or 4th quadrant, where cosine is positive.
This is why when we solve $\sin^{2}y+\cos^{2}y=1$ for $\cos y$,
we take the positive square root:
$\cos y=+\sqrt{1-\sin^{2}y}$
Because of the way we defined y, it is that $\sin^{2}y=(\sin y)^{2}=x^{2}$,
so
$\cos y=+\sqrt{1-x^{2}}\qquad$ or,
$\cos(\sin^{-1}x)=\sqrt{1-x^{2}}$
