Question

cos-1x + cos-1 {x/2 + (root(3-3x2)/2} = pie/3

Answer 1

Given:

$cos^{-1}x+cos^{-1}[\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}]$

Take

$cos^{-1}x=A$

and

$cos^{-1}[\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}]=B$

$\implies\;x=cosA$ and

$\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}=cosB$

$\implies\frac{x+\sqrt{3-3x^2}}{2}=cosB$

$\implies\frac{cosA+\sqrt3\sqrt{1-cos^2A}}{2}=cosB$

$\implies\frac{cosA+\sqrt3\;sinA}{2}=cosB$

$\implies\frac{1}{2}cosA+\frac{\sqrt3}{2}sinA=cosB$

$\implies\;cos\frac{\pi}{3}\;cosA+sin\frac{\pi}{3}\;sinA=cosB$

Using

$\boxed{\bf\;cos(A-B)=cosA\;cosB+sinA\;snB}$

$\implies\;cos(\frac{\pi}{3}-A)=cosB$

$\implies\;\frac{\pi}{3}-A=B$

$\implies\;A+B=\frac{\pi}{3}$

$i.e\boxed{\bf\;cos^{-1}x+cos^{-1}[\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}]=\frac{\pi}{3}}$