Express $\arccos x+ \arccos \left(\frac{x}{2} +\frac{\sqrt{3}-3x^2}{2}\right)$ in simplest form

Express $$\arccos x+ \arccos \left(\frac{x}{2} +\frac{\sqrt{3}-3x^2}{2}\right)$$ in simplest form.

I tried applying the direct formula but, that gave a very complicated equation. Converting it into a trigonometric ratio also seemed complex. How should I try to solve it?

• – Shaun Aug 29 '17 at 15:05
• What is “the direct formula”? – Chase Ryan Taylor Aug 29 '17 at 15:43
• It's arc cos(xy+√(1-x^2)√(1-y^2)) – user473333 Aug 29 '17 at 15:47
• @user473333 You should put that in your question using MathJax markdown. – Chase Ryan Taylor Aug 29 '17 at 16:15

Let $\arccos x=t\implies0\le t\le\pi$ (Check for principal values)
and $x=\cos t,\sin t=+\sqrt{1-x^2}$
$\dfrac x2+\dfrac{\sqrt3\sqrt{1-x^2}}2=\cos\left(t-\dfrac\pi3\right)$
Now $\arccos\left(\dfrac x2+\dfrac{\sqrt3\sqrt{1-x^2}}2\right)=\begin{cases}t-\dfrac\pi3\text{ iff }0\le t-\dfrac\pi3\le\pi\iff\dfrac\pi3\le t\le\pi+\dfrac\pi3\\=-\left(t-\dfrac\pi3\right)\text{ iff }0\le\dfrac\pi3-t\le\pi\iff?\end{cases}$