<s>[INST] How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$? 
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.
 [/INST] 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, $\cos(\sin^{-1}x)=\sqrt{1-x^2}$
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