Prove that: $\;$ $\cos^{-1} \left(2 x^2 - 1\right) = 2 \cos^{-1} \left(x\right), \hspace{1cm} 0 < x <1$
Let $\;$ $x = \cos \theta \implies \theta = \cos^{-1} \left(x\right), \hspace{1cm} \theta \in \left[0, \pi\right]$
$\begin{aligned}
\text{Then, } \cos^{-1} \left(2 x^2 - 1\right) & = \cos^{-1} \left(2 \cos^2 \theta - 1\right) \\\\
& = \cos^{-1} \left[\cos \left(2 \theta\right)\right] \hspace{1cm} \left[\text{Note: } \cos 2 \theta = 2 \cos^2 \theta - 1\right] \\\\
& = 2 \theta \\\\
& = 2 \cos^{-1} \left(x\right)
\end{aligned}$
Hence proved.