Finding exact value of trigonometric functions I was wondering, how do I get the exact fraction (the value) of this trigonometric function: 
$$\cos\left(\sin^{-1}(12/13)+\sin^{-1}(4/5)\right)$$
Usually, I would evaluate the inverse sin in degree mode and multiply (by hand) by $\pi/180$.
But in this case, I don't get exact values of angles
For example :
$$\begin{eqnarray}
\sin^{-1}(12/13)=67,38...^\circ\\
\sin^{-1}(4/5)=53,13...^\circ
\end{eqnarray}
$$
Is there any way of doing it ?
Thank you!

We have:
$$\arcsin\frac{4}{5} = \arg(3+4i),\qquad \arcsin\frac{12}{13}=\arg(5+12i), $$
hence:
$$\arcsin\frac{4}{5}+\arcsin\frac{12}{13}=\arg((3+4i)(5+12i))=\arg(-33+56i)$$
and:
$$ \cos\left(\arcsin\frac{4}{5}+\arcsin\frac{12}{13}\right)=\frac{-33}{\sqrt{33^2+56^2}}=-\frac{33}{65}.$$

Avoiding complex numbers.
$$\arcsin\frac{4}{5} = \arctan\frac{4}{3},\qquad \arcsin\frac{12}{13}=\arctan\frac{12}{5}, $$
so:
$$\arcsin\frac{4}{5}+\arcsin\frac{12}{13}=\arctan\frac{\frac{4}{3}+\frac{12}{5}}{1-\frac{4}{3}\cdot\frac{12}{5}}=\arctan\left(-\frac{56}{33}\right) $$
and:
$$\cos\arctan\left(-\frac{56}{33}\right)=\frac{-33}{\sqrt{33^2+56^2}}=-\frac{33}{56}.$$
