Answer
$$\sin\frac{13\pi}{12}=\frac{\sqrt2-\sqrt6}{4}$$
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 $\frac{\pi}{3}$.
$$\sin\frac{13\pi}{12}$$
First, we separate $13\pi$ into $12\pi$ and $\pi$. In detail,
$$\sin\frac{13\pi}{12}=\sin\Big(\frac{12\pi+\pi}{12}\Big)=\sin\Big(\frac{12\pi}{12}+\frac{\pi}{12}\Big)=\sin\Big(\pi+\frac{\pi}{12}\Big)$$
Now we use the sine sum identity:
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
Therefore,
$$\sin\frac{13\pi}{12}=\sin\pi\cos\frac{\pi}{12}+\cos\pi\sin\frac{\pi}{12}$$
$$\sin\frac{13\pi}{12}=0\times\cos\frac{\pi}{12}+(-1)\times\sin\frac{\pi}{12}$$
$$\sin\frac{13\pi}{12}=-\sin\frac{\pi}{12}$$
Now, we try write $\pi$ as the difference of $4\pi$ and $3\pi$.
$$-\sin\frac{\pi}{12}=-\sin \Big(\frac{4\pi-3\pi}{12}\Big)=-\sin\Big(\frac{4\pi}{12}-\frac{3\pi}{12}\Big)=-\sin\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$$
So, $$\sin\frac{13\pi}{12}=-\sin\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$$
We then use the sine difference identity:
$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
Therefore,
$$\sin\frac{13\pi}{12}=-\Big(\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}\Big)$$
$$\sin\frac{13\pi}{12}=-\Big(\frac{\sqrt3}{2}\frac{\sqrt2}{2}-\frac{1}{2}\frac{\sqrt2}{2}\Big)$$
$$\sin\frac{13\pi}{12}=-\Big(\frac{\sqrt6}{4}-\frac{\sqrt2}{2}\Big)$$
$$\sin\frac{13\pi}{12}=\frac{\sqrt2}{2}-\frac{\sqrt6}{4}$$
$$\sin\frac{13\pi}{12}=\frac{\sqrt2-\sqrt6}{4}$$
