Answer
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\frac{\sqrt2+\sqrt6}{4}$$
Work Step by Step
$$\sin\Big(-\frac{7\pi}{12}\Big)$$
From Negative-Angle Identities: $\sin(-\theta)=-\sin\theta$.
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\sin\frac{7\pi}{12}$$
Next we rewrite $7\pi$ as the sum of $3\pi$ and $4\pi$.
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\sin\frac{3\pi+4\pi}{12}=-\sin\Big(\frac{3\pi}{12}+\frac{4\pi}{12}\Big)=-\sin\Big(\frac{\pi}{4}+\frac{\pi}{3}\Big)$$
Then we apply the identity for sum of sines:
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
with $A=\frac{\pi}{4}$ and $B=\frac{\pi}{3}$
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\Big(\sin\frac{\pi}{4}\cos\frac{\pi}{3}+\cos\frac{\pi}{4}\sin\frac{\pi}{3}\Big)$$
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\Big(\frac{\sqrt2}{2}\frac{1}{2}+\frac{\sqrt2}{2}\frac{\sqrt3}{2}\Big)$$
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\Big(\frac{\sqrt2}{4}+\frac{\sqrt6}{4}\Big)$$
$$\sin\Big(-\frac{7\pi}{12}\Big)=-\frac{\sqrt2+\sqrt6}{4}$$
