Question

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

Answer 1

$3cos^{-1x}=cos^{-1}(4x^3-3x)\,x\in[-\frac{1}{2},1]$
To prove:
Let x = cosθ. Then, $cos^{-1}x=\theta$.
We have,
RHS= $cos^{-1}(4x^3-3x)$
= $cos^{-1}(4cos2\theta-3cos\theta)$
= $cos^{-1}(cos3\theta)$
= 3θ
= $3cos^{-1x}$
$=L.H.S.$