Question

If $\frac{1}{2}\le x\le 1,$ then ${\cos }^{-1}x+{\cos }^{-1}(\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2})=$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\pi$
• D. $0$

Answer 1

Answer: (B)

$\frac{\pi }{3}$

Let $x=\cos \theta$

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

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

$=[\cos \theta \cos 60°+\sin \theta \sin 60°]$

$={\cos }^{-1}\left\{\cos [60°-\theta ]\right\}$

$=[{60}^o-\theta ]$

Then,

${\cos }^{-1}x+{\cos }^{-1}\left\{\cos (60°-\theta )\right\}$

${\cos }^{-1}\left(\cos \theta \right)+{\cos }^{-1}\left\{\cos (60°-\theta )\right\}$

$=\theta +60°-\theta$

$=60°=\frac{\pi }{3}$