Answer
$$\cos(-15^\circ)=\frac{\sqrt 6+\sqrt 2}{4}$$
Work Step by Step
$$A=\cos(-15^\circ)$$
We can rewrite $-15^\circ$ into the difference of $30^\circ$ and $45^\circ$, which means $$A=\cos(30^\circ-45^\circ)$$
Now we apply the identity for cosine of a difference: $$A=\cos30^\circ\cos45^\circ+\sin30^\circ\sin45^\circ$$ $$A=\frac{\sqrt 3}{2}\frac{\sqrt 2}{2}+\frac{1}{2}\frac{\sqrt 2}{2}$$ $$A=\frac{\sqrt 6}{4}+\frac{\sqrt 2}{4}$$ $$A=\frac{\sqrt 6+\sqrt 2}{4}$$