## Calculus: Early Transcendentals 8th Edition

let $\sin^{-1}x=\theta, -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$ $\sin\theta=x$ $\because \sin^2\theta+\cos^2\theta=1$ $\therefore \cos(\sin^{-1}x)=\cos\theta=\pm\sqrt{1-\sin^2\theta}=\pm\sqrt{1-x^2}$ Since $\cos\theta\geq0$ for $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2} ,$ $\therefore \cos(\sin^{-1}x)=\sqrt{1-x^2}$
$\cos(\sin^{-1}x)=\sqrt{1-x^2}$ is proven above.