Answer
$cos~15^{\circ} = \frac{\sqrt{6}+\sqrt{2}}{4} = 0.9659$
Work Step by Step
In the triangle $EAD$, the length of $AD$ is $\sqrt{6}+\sqrt{2}$
In the triangle $EAD$, the length of $ED$ is twice the radius, so the length of $ED$ is $4$
Note that triangle $EAD$ is a right triangle since the angle $EAD = 90^{\circ}$. We can use angle $ADB$ of triangle $EAD$ to find $cos~15^{\circ}$:
$cos~15^{\circ} = \frac{adjacent}{hypotenuse}$
$cos~15^{\circ} = \frac{AD}{ED}$
$cos~15^{\circ} = \frac{\sqrt{6}+\sqrt{2}}{4} = 0.9659$
