Answer
$\dfrac{\sqrt6+\sqrt2}{4}$
Work Step by Step
Recall:
$\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$
Hence,
$\cos{15^\circ}\\
=\cos{(45^\circ-30^\circ)}\\
=\cos{45^\circ}\cos{30^\circ}+\sin{45^\circ}\sin{30^\circ}\\
=\frac{\sqrt3}{2}\frac{\sqrt2}{2}+\frac{1}{2}\frac{\sqrt2}{2}\\
=\frac{\sqrt6+\sqrt2}{4}$
