Question

Prove that if $\frac{1}{2}\le x\le 1$ then, $co{s}^{-1}x+co{s}^{-1}[\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}]=\frac{\pi }{3}.$

Answer 1

Let $\theta =co{s}^{-1}x.$ Then for all $ϵ[\frac{1}{2},1],\theta ϵ[0,\frac{\pi }{3}].$ Also $x=cos\theta .$
Consider LHS : Let $y=co{s}^{-1}x+co{s}^{-1}[\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}]=\theta +co{s}^{-1}\left[\frac{cos\theta }{2}\right]+\frac{\sqrt{3}}{2}\sqrt{1-co{s}^2\theta }$
$\Rightarrow y=\theta +co{s}^{-1}[\frac{1}{2}cos\theta +\frac{\sqrt{3}}{2}sin\theta ]=\theta +co{s}^{-1}\left[cos\right(\theta -\frac{\pi }{3}\left)\right]$
$\because 0\le \theta \le \frac{\pi }{3}\Rightarrow -\frac{\pi }{3}\le \theta -\frac{\pi }{3}\le 0$ $\Rightarrow 0\le \frac{\pi }{3}-\theta \le \frac{\pi }{3}$
$\therefore y=\theta +co{s}^{-1}\left[cos\right(\frac{\pi }{3}-\theta \left)\right]=\theta +\frac{\pi }{3}-\theta =\frac{\pi }{3}=RHS.$