To find $\cos 15^\circ$, we can use the angle subtraction formula:

$\cos(a - b) = \cos a \cos b + \sin a \sin b$

For $a = 45^\circ$ and $b = 30^\circ$:

$\cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 45^\circ = \frac{\sqrt{2}}{2}$ $\cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}$

Substituting these into the formula:

$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$

$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$

$= \frac{\sqrt{6} + \sqrt{2}}{4}$

So the correct answer is A) $\frac{\sqrt{6} + \sqrt{2}}{4}$.