Question

The value of $co{s}^{-1}x+co{s}^{-1}(\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2})$ $(1/2\le x\le 1)$
is equal to

• A. $\pi /6$
• B. $\pi /3$
• C. $\pi$
• D. 0

Answer 1

Answer: (B)

$\pi /3$

Let $co{s}^{-1}x=y$. Then $x=cosy$, so that $1/2\le x\le 1$, or $0\le y\le x/3$, and
$\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}=\frac{1}{2}cosy+\frac{\sqrt{3}}{2}siny$
$=cos\frac{\pi }{3}cosy+sin\frac{\pi }{3}siny=cos(\frac{\pi }{3}-y)$
$\Rightarrow co{s}^{-1}(\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2})=\frac{\pi }{3}-y$
because $co{s}^{-1}\left(cosx\right)=x$ if $0\le x\le \pi$. Therefore, the given expression is equal to $y+\pi /3-y=\pi /3$