\[ cos^{-1}(x)=\theta \] then \[ x=cos\theta \]................................................(1)
Case 1 \[0\leqslant x\leq 1\]
\[0\leqslant x\leq 1\]
\[0 \leq cos\theta \leq 1\]
\[0 \leq \theta \leq \frac{\Pi }{2}\]
\[0 \leq 2\theta \leq \Pi \]
Also \[0 \leq x \leq 1\]
\[-1\leq 2x^2-1 \leq 1\]
\[cos2\theta =2x^2-1 \]
\[2\theta = cos^{-1}(2x^2-1 )\]
\[2cos^{-1}(x) = cos^{-1}(2x^2-1 )\]
Case 2 \[ -1\leq x\leq 0\]
That implies \[ -1\leq cos\theta \leq 0\] from equation 1
\[\frac{\Pi }{2}\leq \theta \le\Pi \]
\[\Pi \leq 2\theta \le2\Pi \]
\[-2\Pi \leq -2\theta \le-\Pi \]
\[0 \leq2\pi -2\theta \le\pi \]
Also \[ -1\leq x\leq 0\]
hence \[ -1\leq 2x^2-1\leq1\]
\[cos2\theta= 2x^2-1\]
\[cos(2\pi-2\theta)= 2x^2-1\]
\[(2\pi-2\theta)= cos^{-1}(2x^2-1)\]
\[ 2\pi-2cos^{-1}(x)=cos^{-1}(2x^2-1)\]
Therefore \[2cos^{-1}(x)=2\pi-cos^{-1}(2x^2-1)\]
hence proved
