Question

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

Answer 1

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

Thus we have,

$\text{R.H.S.}={\cos }^{-1}({4x}^3-3x)$

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

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

$=3\theta =3{\cos }^{-1}x=\text{L.H.S.}$