Question

Prove that $3{\cos }^{-1}x={\cos }^{-1}(4x^3-3x),xϵ\left[\begin{array}{c}\frac{1}{2},1\end{array}\right]$.

Answer 1

We have to prove that $3{\cos }^{-1}x={\cos }^{-1}(4x^3-3x)$

First consider RHS, ${\cos }^{-1}(4x^3-3x)$

Let us take $x=cos\theta$, we know that $cos\left(3\theta \right)=4co{s}^3\theta -3cos\theta$

$⟹{\cos }^{-1}(4x^3-3x)={\cos }^{-1}(4{cos}^3\theta -3cos\theta )={\cos }^{-1}\left(cos\right(3\theta \left)\right)=3\theta$

$\because \theta =co{s}^{-1}x$, $⟹{\cos }^{-1}(4x^3-3x)=3co{s}^{-1}x$

$\therefore LHS=RHS$

$0\le 3co{s}^{-1}x\le \pi$

$0\le {\cos }^{1}x\le \pi /3$

$\therefore xϵ[\frac{1}{2},1]$

Hence proved.