Question

Prove that :
${\cos }^{-1}\left(x\right)+{\cos }^{-1}\{\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}\}=\frac{\pi }{3}$

Answer 1

Let $x=\cos y⟹y={\text{cos}}^{-1}x$

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

$={\text{cos}}^{-1}\{\frac{\cos y}{2}+\frac{\sqrt{3(1-{\cos }^{2}y)}}{2}\}$

$={\text{cos}}^{-1}\{\frac{1}{2}\cos y+\frac{\sqrt{3}}{2}\sin y\}$

$={\text{cos}}^{-1}\{\cos \frac{\pi }{3}\cos y+\sin \frac{\pi }{3}\sin y\}$

$={\text{cos}}^{-1}\{\cos (\frac{\pi }{3}-y)\}=\frac{\pi }{3}-y$

${\text{cos}}^{-1}\left(x\right)+{\text{cos}}^{-1}\{\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}\}=y+\frac{\pi }{3}-y=\frac{\pi }{3}$

Hence Proved