# Inverse trigonometry?

• May 2nd 2009, 10:26 PM
fardeen_gen
Inverse trigonometry?
Prove that:
$2\arctan\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] = \arccos\left[\frac{b + a\cos x}{a + b\cos x}\right]$
• May 3rd 2009, 12:15 AM
Infophile
We can differentiate each term and show that $f'(x)=g'(x)$ for all $x$.

Thus $f(x)=g(x)+C$ and taking $x=0$ give $C=0$.

There are many calculations but it isn't difficult for all that.

:)