Answer
$$\sin\Big(-\frac{7\pi}{12}\Big)=\frac{-\sqrt6-\sqrt2}{4}$$
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 $\frac{\pi}{3}$.
$$\sin\Big(-\frac{7\pi}{12}\Big)$$
From Negative-Angle Identities:
$$\sin(-\theta)=-\sin\theta$$
So that means, $$\sin\Big(-\frac{7\pi}{12}\Big)=-\sin\frac{7\pi}{12}$$
Now, we write $7\pi$ as the sum of $4\pi$ and $3\pi$.
$$\sin\frac{7\pi}{12}=\sin \Big(\frac{4\pi+3\pi}{12}\Big)=\sin\Big(\frac{4\pi}{12}+\frac{3\pi}{12}\Big)=\sin\Big(\frac{\pi}{3}+\frac{\pi}{4}\Big)$$
Therefore, $$\sin\Big(-\frac{7\pi}{12}\Big)=-\sin\Big(\frac{\pi}{3}+\frac{\pi}{4}\Big)$$
We then use the sine sum identity:
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
Therefore,
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\Big(\sin\frac{\pi}{3}\cos\frac{\pi}{4}+\cos\frac{\pi}{3}\sin\frac{\pi}{4}\Big)$$
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\Big(\frac{\sqrt3}{2}\frac{\sqrt2}{2}+\frac{1}{2}\frac{\sqrt2}{2}\Big)$$
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\Big(\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\Big)$$
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\frac{\sqrt6+\sqrt2}{4}$$
$$\sin\Big(-\frac{7\pi}{12}\Big)=\frac{-\sqrt6-\sqrt2}{4}$$
