### Instruction: Compute $\cos 15^\circ.$.### Response: One way to approach this problem is to use the trigonometric identity $\cos (A - B) = \cos A \cos B + \sin A \sin B$ and apply it to $A = 45^\circ$ and $B = 30^\circ.$
This gives us $\cos 15^\circ = \cos (45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ.$
Now we need to recall the values of $\cos 45^\circ, \cos 30^\circ, \sin 45^\circ,$ and $\sin 30^\circ.$
These are common values that can be derived from right triangles with special angles, or by using the unit circle.
We have $\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}, \cos 30^\circ = \frac{\sqrt{3}}{2},$ and $\sin 30^\circ = \frac{1}{2}.$
Substituting these values into the equation, we get $\cos 15^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}.$
Simplifying the fractions, we get $\cos 15^\circ = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}.$
Combining the terms, we get $\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}.$