Answer
$$-\cos15^\circ=\frac{-\sqrt2-\sqrt6}{4}$$
Work Step by Step
Recall that when we need to calculate cosine of unfamiliar angles, we try to rewrite them in terms of familiar angles, which are $30^\circ$, $45^\circ$ and $60^\circ$.
Therefore, we rewrite $15^\circ$ as follows:
$$15^\circ=60^\circ-45^\circ$$
That means $$\cos15^\circ=\cos(60^\circ-\cos45^\circ)$$
Now note that $$\cos(A-B)=\cos A\cos B+\sin A\sin B\hspace{1cm}\text{(cosine difference identity)}$$
So, $$\cos15^\circ=\cos60^\circ\cos45^\circ+\sin60^\circ\sin45^\circ$$
$$\cos15^\circ=\frac{1}{2}\frac{\sqrt2}{2}+\frac{\sqrt3}{2}\frac{\sqrt2}{2}$$
$$\cos15^\circ=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}$$
$$\cos15^\circ=\frac{\sqrt2+\sqrt6}{4}$$
Therefore, $$-\cos15^\circ=\frac{-\sqrt2-\sqrt6}{4}$$
