Find the imaginary part of \[(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6.\]
Using the sum-to-product formula, we have
\begin{align*}
\cos 12^{\circ} + \cos 48^{\circ} &= 2 \cos 30^{\circ} \cos 18^{\circ}, \\
\sin 12^{\circ} + \sin 48^{\circ} &= 2 \sin 30^{\circ} \cos 18^{\circ}. \\
\end{align*}Therefore,
\begin{align*}
(\cos 12^{\circ} + i \sin 12^{\circ} + \cos 48^{\circ} + i \sin 48^{\circ})^6 &= [2\cos 18^{\circ} (\cos 30^{\circ} + i \sin 30^{\circ})]^6 \\
&= 64\cos^6 18^{\circ} (\cos 180^{\circ} + i \sin 180^{\circ}) \\
&= -64\cos^6 18^{\circ},
\end{align*}which is real. Hence, the imaginary part is $\boxed{0}$.

(See if you can also find a geometric solution!)