Using Euler's formula for cos/sin 12 degrees and cos/sin 48 degrees $(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)$. Using Euler's Formula, turn this into exponential form (i.e. something like $e^{i\frac{5\pi}{12}}$). 
Would I need to use the $\cos$ and $\sin$ sum and difference formulas? I tried doing that and it became messy very quickly. Is there an alternative? 

Observe
\begin{align*}
\cos 12^{\circ}=\cos(30^{\circ}-18^{\circ})&=\cos30^{\circ}\cos18^{\circ}+\sin30^{\circ}\sin18^{\circ} \\
\sin 12^{\circ}=\sin(30^{\circ}-18^{\circ})&=\sin30^{\circ}\cos18^{\circ}-\cos30^{\circ}\sin18^{\circ}\\
\cos 48^{\circ}=\cos(30^{\circ}+18^{\circ})&=\cos30^{\circ}\cos18^{\circ}-\sin30^{\circ}\sin18^{\circ} \\
\sin 48^{\circ}=\sin(30^{\circ}+18^{\circ})&=\sin30^{\circ}\cos18^{\circ}+\cos30^{\circ}\sin18^{\circ}
\end{align*}
Then, from trigonometric identities, we can reduce it as follows
\begin{align*}
\cos12^{\circ}+i\sin12^{\circ}+\cos48^{\circ}+i\sin48^{\circ}&=2\cos30^{\circ}\cos18^{\circ}+i2\sin30^{\circ}\cos18^{\circ}\\
&=2e^{i\frac{\pi}{6}}\cos\left(\frac{\pi}{10}\right)
\end{align*}
