How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$?

Question

How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$?

I remember having to draw a triangle, but I'm not sure anymore.

Answer 1

Given $\cos(\sin^{-1}x)$

Let $\sin^{-1}x=\theta$ $$\implies\sin\theta=x\\\sin^2\theta=x^2\\1-\cos^2\theta=x^2\\\cos^2\theta=1-x^2\\\cos\theta=\pm\sqrt{1-x^2}\\\theta=\cos^{-1}\pm\sqrt{1-x^2}$$ Now, plug in $\theta=\cos^{-1}\pm\sqrt{1-x^2}$ in $\cos(\sin^{-1}x)$ and we get$$\pm\sqrt{1-x^2}$$But note that $\sin^{-1}x$ is in $\left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]$ so $\cos(\sin^{-1}x)\ge0$

Therefore, $\cos(\sin^{-1}x)=\sqrt{1-x^2}$

Answer 2

$\cos(\sin^{-1}(x)) = \cos(\sin^{-1}(x/1))$

Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $\sqrt{1^2 - x^2}$