### Instruction: Compute $\cos 15^\circ.$.### Response: I recall that there is a formula for $\cos (A \pm B)$ in terms of $\cos A, \cos B, \sin A,$ and $\sin B.$
Maybe I can use that to simplify $\cos 15^\circ$ by finding a way to write $15^\circ$ as a sum or difference of two angles whose cosines and sines I know.
One possible choice is $15^\circ = 45^\circ - 30^\circ,$ since I know the values of $\cos 45^\circ, \cos 30^\circ, \sin 45^\circ,$ and $\sin 30^\circ.$
So, using the formula, I get $\cos 15^\circ = \cos (45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ.$
Now I plug in the values that I know: $\cos 45^\circ = \frac{\sqrt{2}}{2}, \cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 45^\circ = \frac{\sqrt{2}}{2}, \sin 30^\circ = \frac{1}{2}.$
This gives me $\cos 15^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}.$