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Q.
The value of $\cos^{-1} x + \cos^{-1} \left( \frac{x}{2} + \frac{1}{2} \sqrt{3 - 3x^2} \right) ; \frac{1}{2} \le
x \le 1$ is
Inverse Trigonometric Functions
Solution:
Let $\cos^{-1} x = y$
$ \Rightarrow x = \cos y, $ so that $\frac{1}{2} \le x \le 1$
or $ 0 \le y \le \frac{\pi}{3} $
and $\frac{x}{2} + \frac{1}{2} \sqrt{3 - 3x^{2}} $
$= \frac{1}{2} \cos y + \frac{\sqrt{3}}{2} \sin y$
$ = \cos \frac{\pi}{3} \cos y + \sin \frac{\pi}{3} \sin y = \cos\left(\frac{\pi}{3} - y\right) $
$\Rightarrow \cos^{-1} \left( \frac{x}{2} + \frac{1}{2} \sqrt{3 -3x^{2}}\right) = \frac{\pi}{3} -y$
$ \therefore $ the given expression is equal to
$y + \frac{\pi}{3} - y = \frac{ \pi}{3}$