Answer
$ \dfrac{\sqrt {3}}{2}$
Work Step by Step
Recall the rule: $( \ g \circ \ p ) =g[p(x)]$
Let us consider that $g (x)=\cos x $ and $p(x)= \dfrac{x}{2}$
Now, $( g \ \circ \ p ) (60^{\circ}) =g[p(60)^{\circ}] \\=
g (\dfrac{60^{\circ}}{2}) \\=g ( 30^{\circ}) $
We know from the unit circle that $\cos (30^{\circ}) =\dfrac{\sqrt {3}}{2}$
Thus, we have: $( g \ \circ \ p ) (60^{\circ}) =\cos (30^{\circ}) = \dfrac{\sqrt {3}}{2}$
