Prove $3 \cos ^{-1} x = \cos ^{-1} (4 x^{3} - 3 x)$,where $x \in [\frac{1}{2}, 1]$.

To prove $3 \cos ^{-1} x = \cos ^{-1} (4 x^{3} - 3 x)$ for $x \in [\frac{1}{2}, 1]$.
Let $x = \cos \theta$. Then $\theta = \cos ^{-1} x$.
Since $x \in [\frac{1}{2}, 1]$,we have $\cos \theta \in [\frac{1}{2}, 1]$,which implies $\theta \in [0, \frac{\pi}{3}]$.
Multiplying by $3$,we get $3\theta \in [0, \pi]$.
Now,consider the $R.H.S$:
$\cos ^{-1} (4 x^{3} - 3 x)$
$= \cos ^{-1} (4 \cos ^{3} \theta - 3 \cos \theta)$
Using the trigonometric identity $\cos 3\theta = 4 \cos ^{3} \theta - 3 \cos \theta$,we get:
$= \cos ^{-1} (\cos 3\theta)$
Since $3\theta \in [0, \pi]$,we have $\cos ^{-1} (\cos 3\theta) = 3\theta$.
$= 3 \cos ^{-1} x = L.H.S$.
Hence,the identity is proved.