Question

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

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

Answer 1

The correct option is B $\frac{\pi }{3}$

Let $co{s}^{-1}x=y.$ Then $x=cosy$, such that $\frac{1}{2}\le x\le 1,or0\le y\le \frac{\pi }{3}$, 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 $y+\frac{\pi }{3}-y=\frac{\pi }{3}$