why does $\sin(2\sin^{-1}{(x/2)}) = x\sqrt{1-x^2/2}$

Question

$$\sin\left(2\sin^{-1}{\left(\frac{x}{2}\right)}\right) = x\sqrt{1-\frac{x^2}{2}}$$

I don't understand how to prove this statement.

Answer 1

$$\sin(2u)=2\sin(u)\cos(u)=2\sin(u)\sqrt{1-\sin^2(u)}$$Let $u=\sin^{-1}\left(\frac{x}{2}\right)$ $$\sin\left(2\sin^{-1}\left(\frac{x}{2}\right)\right)=2\left(\frac{x}{2}\right)\sqrt{1-\left(\frac{x}{2}\right)^2}$$

Answer 2

$$ \begin{split} \sin(2\sin^{-1}(x/2))&=2\sin(\sin^{-1}(x/2))\cos(\sin^{-1}(x/2))\\ &=x\cos(\sin^{-1}(x/2))\\ &=x\sqrt{1-\sin(\sin^{-1}(x/2))^2}\\ &=x\sqrt{1-\frac{x^2}4} \end{split} $$