Hint:

use: $$ \cos 2 \alpha= \cos^2 \alpha-\sin^2 \alpha= 2\cos^2 \alpha-1 $$ for $\alpha=\cos^{-1} x$. Than note that , by definition, $\cos (\cos^{-1} x)=x$

So you find: $$ \cos (2 \cos^{-1} x)=2\cos^2(\cos^{-1} x)-1=2x^2-1 $$

This is $\cos\Big(2\times\text{something}\Big)$. There's a standard identity for that: \begin{align} \cos\Big(2\times\text{something}\Big) & = \cos^2\Big(\text{something}\Big) - \sin^2\Big(\text{something}\Big) \\[10pt] & = \Big(\cos(\text{something})\Big)^2 - \Big(\sin^2(\text{something})\Big)^2 \end{align} Then you need to deal with $\cos(\arccos(x))$ (that part is easy) and $\sin(\arccos(x))$. The latter can be done via a diagram like the one you've drawn.