Let f : [1/2, 1] → [-1, 1] is given by f^-1(x) = 4x^3− 3x

$y=4x^3-3x$  let $x=\cos θ$

$y=4\cos^3 θ-3\cos θ$ 

$⇒θ\cos^{-1}(x)$

$y=\cos 3θ⇒\frac{1}{3}\cos^{-1}y=θ$

so $\cos\left(\frac{1}{3}\cos^{-1}y\right)=x$

$f^{-1}x=\cos \left(\frac{1}{3} \cos ^{-1} x\right)$