Answer
$\dfrac{\sqrt 6+\sqrt 2}{4}$
Work Step by Step
We know that $ \cos (x-y)= \cos x \cos y+ \sin x \sin y$
$\cos 15^{\circ}=\cos (45^{\circ}-30^{\circ})=\cos 45^{\circ} \cos 30^{\circ}+ \sin 45^{\circ} \sin 30^{\circ}$
$=(\dfrac{\sqrt 2}{2})(\dfrac{\sqrt 3}{2})+(\dfrac{\sqrt 2}{2}) (\dfrac{1}{2})$
$=\dfrac{\sqrt 6+\sqrt 2}{4}$
