Prove $\cos^{-1}(x) = 2\sin^{-1} \sqrt{ \frac{1-x} {2} }$ for $-1 \leq x \leq 1$

Question

I'm attempting to prove this using inverse function theorems but unsure where to start.

Answer 1

It is usually best to start from the more complicated side and see how things can simplify. Here it ends up with the double-angle formulae for $\cos y$. $$\begin{align}&y=2\sin^{-1}\sqrt{\frac{1-x}2}\\ \implies&\sin\frac y2=\sqrt{\frac{1-x}2}\\ \implies& 1-2\sin^2\frac y2=x\\ \implies&\cos y=x\\ \implies& y=\cos ^{-1}x\end{align}$$

Answer 2

Hint: Put $x=\cos(2t)$, then $\displaystyle\frac{1-x}{2}=\frac{1-\cos(2t)}{2}=\sin^{2}(t)$