Question

Question #13738

Answer 1

`\sin(2\sin^{-1}(x))=2x\sqrt{1-x^2}`.

Explanation:

This requires a little care. The best approach is to define `\theta=\sin^{-1}(x)` so that \sin \theta=x#. Then we can put in the double angle formula

`\sin(2\theta)=2\sin \theta \cos \theta`

We have `\sin \theta=x` and from the Pythagorean identity, `\cos \theta=\pm\sqrt{1-x^2}`. But also, for the inverse sine function, `-(\pi/2)\le \theta \le (\pi/2)` so `\cos \theta \ge 0`. So then

`\sin(2\theta)=2\sin \theta \cos \theta=+2(x)(\sqrt{1-x^2})`.