<s>[INST] Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$ A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.
By taking $\cos\alpha  = x$, the equation on the right can be simplified to $\cos^{-1}\left(\cos\left(\frac {\pi}{3} - \alpha\right)\right)$. Finally, you get $\frac{\pi}{3}$ as the final answer.
But the answers are $\frac{\pi}{3}$ and  $2\cos^{-1}\left(\frac{1}{3}\right)-\frac{\pi}{3}$. How does it work out that way?
 [/INST] Using Principal values
$$-\dfrac\pi3\le\cos^{-1}x-\dfrac\pi3\le\pi-\dfrac\pi3 $$
$$\cos^{-1}x-\dfrac\pi3=\begin{cases} \cos^{-1}\left(\dfrac x2+\dfrac{\sqrt{3(1-x^2)}}2\right) &\mbox{if } \cos^{-1}x-\dfrac\pi3\ge0\iff  x\le\cos\dfrac\pi3 \\
-\cos^{-1}\left(\dfrac x2+\dfrac{\sqrt{3(1-x^2)}}2\right) & \mbox{if }   x>\cos\dfrac\pi3\end{cases}$$
Now observe that $\dfrac13<\cos\dfrac\pi3=\dfrac12<\dfrac23$
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