Compute:
\[
\sin \frac{\pi}{12} + \sin \frac{3\pi}{12} + \sin \frac{5\pi}{12} + \sin \frac{7\pi}{12} + \sin \frac{9\pi}{12} + \sin \frac{11\pi}{12}.
\]
Notice that problem has three pairs of the form $\sin \theta + \sin(\pi - \theta).$  The sum-to-product formula yields
\begin{align*}
\sin \frac{\pi}{12} + \sin \frac{11\pi}{12} &= 2 \sin \frac{\pi}{2} \cos \frac{5\pi}{12} \\
&= 2 \cos \frac{5\pi}{12}, \\
\sin \frac{3\pi}{12} + \sin \frac{9\pi}{12} &= 2 \sin \frac{\pi}{2} \cos \frac{\pi}{4} \\
&= \sqrt{2}, \\
\sin \frac{5\pi}{12} + \sin \frac{7\pi}{12} &= 2 \sin \frac{\pi}{2} \cos \frac{\pi}{12} \\
&= 2 \cos \frac{\pi}{12}.
\end{align*}Applying the sum-to-product formula once more yields
\begin{align*}
& \sin \frac{\pi}{12} + \sin \frac{3\pi}{12} + \sin \frac{5\pi}{12} + \sin \frac{7\pi}{12} + \sin \frac{9\pi}{12} + \sin \frac{11\pi}{12} \\
&= \sqrt{2} + 2 \Big(\cos \frac{5\pi}{12} + \cos \frac{\pi}{12} \Big) \\
&= \sqrt{2} + 4 \cos \frac{\pi}{4} \cos \frac{\pi}{6} \\
&= \sqrt{2} + 4 \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} \\
&= \boxed{\sqrt{2} + \sqrt{6}}.
\end{align*}