To Prove:

$3\cos^{-1}x = \cos^{-1}(4x^3 - 3x), \quad \text{where } x \in \left[\frac{1}{2}, 1\right]$

Proof:

Let $x = \cos \theta$. Then $\cos^{-1}x = \theta$.

Since $x \in \left[\frac{1}{2}, 1\right]$, we have $\cos \theta \in \left[\frac{1}{2}, 1\right]$.

This means:

$\theta \in \left[0, \frac{\pi}{3}\right]$

LHS:

$3\cos^{-1}x = 3\theta$

RHS:

$\cos^{-1}(4x^3 - 3x)$

$= \cos^{-1}(4\cos^3\theta - 3\cos\theta)$

Using the identity $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$:

$= \cos^{-1}(\cos 3\theta)$

Since $\theta \in \left[0, \frac{\pi}{3}\right]$, we have:

$3\theta \in [0, \pi]$

This lies within the principal range of $\cos^{-1}$, which is $[0, \pi]$.

Therefore:

$\cos^{-1}(\cos 3\theta) = 3\theta$

Hence, LHS = RHS

$\therefore \; 3\cos^{-1}x = \cos^{-1}(4x^3 - 3x) \text{ is proved.}$