Answer
$$\sin\frac{\pi}{12}=\frac{\sqrt6-\sqrt2}{4}$$
Work Step by Step
*STRATEGY: In this type of exercise, we would try to rewrite the angle in terms of already familiar angles, including $\frac{\pi}{6}$, $\frac{\pi}{4}$ and $\frac{\pi}{3}$.
$$\sin\frac{\pi}{12}$$
Now, we try writing $\pi$ as the difference of $4\pi$ and $3\pi$.
$$\sin\frac{\pi}{12}=\sin \Big(\frac{4\pi-3\pi}{12}\Big)=\sin\Big(\frac{4\pi}{12}-\frac{3\pi}{12}\Big)=\sin\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$$
We then use the sine difference identity:
$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
Therefore,
$$\sin\frac{\pi}{12}=\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}$$
$$\sin\frac{\pi}{12}=\frac{\sqrt3}{2}\frac{\sqrt2}{2}-\frac{1}{2}\frac{\sqrt2}{2}$$
$$\sin\frac{\pi}{12}=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$$
$$\sin\frac{\pi}{12}=\frac{\sqrt6-\sqrt2}{4}$$