Answer
$$\sin(-105^\circ)=\frac{-\sqrt2-\sqrt6}{4}$$
5 would be matched with B.
Work Step by Step
$$\sin(-105^\circ)$$
Remember the Negative-angle Identities:
$$\sin(-\theta)=-\sin\theta$$
So, $$\sin(-105^\circ)=-\sin105^\circ$$
Now, we rewrite $105^\circ$ as the sum of $45^\circ$ and $60^\circ$.
$$105^\circ=45^\circ+60^\circ$$
Therefore, $$-\sin105^\circ=-\sin(45^\circ+60^\circ)$$
Now we apply the sine sum identity, which states
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
That makes
$$-\sin105^\circ=-(\sin 45^\circ\cos60^\circ+\cos45^\circ\sin60^\circ)$$
$$-\sin105^\circ=-\Big(\frac{\sqrt2}{2}\frac{1}{2}+\frac{\sqrt2}{2}\frac{\sqrt3}{2}\Big)$$
$$-\sin105^\circ=-\Big(\frac{\sqrt2}{4}+\frac{\sqrt6}{4}\Big)$$
$$-\sin105^\circ=\frac{-\sqrt2-\sqrt6}{4}$$
Therefore, $$\sin(-105^\circ)=\frac{-\sqrt2-\sqrt6}{4}$$
5 would be matched with B.
