How to show the relation of the Inverse trigonometric functions

Question

I want to show the relation;

$$ \arccos[-1+\frac{Q^2}{2}]=2\arccos[\frac{Q}{2}]$$

Q is a real number. I used Mathematica 8 and I checked this relation is correct. But I don't know how to prove it. I think we have to know the trigonometric addition formulas of arccos.

Would you give me some advice??

Answer 1

Taking the cosine of both members,

$$-1+\frac{Q^2}2=2\left(\frac Q2\right)^2-1.$$

But for the initial identity to be possible, the right arc cosine must be limited to the range $\left[0,\dfrac\pi2\right]$ and its argument must be non-negative. Hence the equality only holds when

$$0\le Q\le2.$$

Answer 2

Let $\arccos\frac Q2=u\implies Q=2\cos u$ and $0\le u\le\pi$

$\arccos\left(-1+\frac{Q^2}2\right)=\arccos(\cos2u)=2u$ if $0\le2u\le\pi$

$=2\pi -2u$ if $\pi <2u\le2\pi$