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

$\dfrac{\sqrt 3}{2}$
We are given that $f(\theta)=\sin \theta$ and $\theta=60^{\circ}$ We know form the unit circle that $\sin (60^{\circ})=\sin (\dfrac{\pi}{3})=\dfrac{\sqrt 3}{2}$ Thus, we have: $f(\theta)=\sin (60^{\circ}) \\=\sin (\dfrac{\pi}{3}) \\=\dfrac{\sqrt 3}{2}$