## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{1}{4}$
Let us consider: $g(\theta)=\cos \theta$ We are given that $\theta=60^{\circ}$ We know form the unit circle that $\cos (60^{\circ})=\cos (\dfrac{\pi}{3})=\dfrac{1}{2}$ Thus, we have: $[g(\theta)]^2=[cos(60^{\circ})]^2 \\=[cos\dfrac{\pi}{3}]^2 \\ =[\dfrac{1}{2}]^2 \\=\dfrac{1}{4}$