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!
 A: 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:
$$\cos\arctan\left(-\frac{56}{33}\right)=\frac{-33}{\sqrt{33^2+56^2}}=-\frac{33}{56}.$$
A: HINT:
Let $A=\arcsin\dfrac{12}{13}$
Using the definition of Principal value of $\arcsin,0<A<\dfrac\pi2$
$\implies \cos A=+\sqrt{1-\sin^2A}=\dfrac5{13}$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\cos\pars{\arcsin\pars{12 \over 13} + \arcsin\pars{4 \over 5}}}
\\[5mm]&=\cos\pars{\arcsin\pars{12 \over 13}}\cos\pars{\arcsin\pars{4 \over 5}}
-\
\overbrace{\sin\pars{\arcsin\pars{12 \over 13}}}^{\dsc{12 \over 13}}\
\overbrace{\sin\pars{\arcsin\pars{4 \over 5}}}^{\dsc{4 \over 5}}
\\[5mm]&=\root{1 - \sin^{2}\pars{\arcsin\pars{12 \over 13}}}
\root{1 - \sin^{2}\pars{\arcsin\pars{4 \over 5}}}\ -\ {12 \over 13}\,{4 \over 5}
\\[5mm]&=\root{1 - \pars{12 \over 13}^{2}}
\root{1 - \pars{4 \over 5}^{2}} - {48 \over 65}
={3 \over 13} - {48 \over 65}=\color{#66f}{\large -\,{33 \over 65}}
\end{align}
