Prove that:

$\displaystyle 2\arctan\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] = \arccos\left[\frac{b + a\cos x}{a + b\cos x}\right]$

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- May 2nd 2009, 09:26 PMfardeen_genInverse trigonometry?
Prove that:

$\displaystyle 2\arctan\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] = \arccos\left[\frac{b + a\cos x}{a + b\cos x}\right]$ - May 2nd 2009, 11:15 PMInfophile
We can differentiate each term and show that $\displaystyle f'(x)=g'(x)$ for all $\displaystyle x$.

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

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

:)