## Precalculus (6th Edition) Blitzer

$\frac{1}{2}$ , $\frac{\sqrt{3}}{2}$ , $\frac{\sqrt{3}}{2}$ , and $\frac{1}{2}$.
By putting the values according to the trigonometric functions, we get: $\sin {{30}^{\circ }}=\frac{1}{2}$ $\cos {{30}^{\circ }}=\frac{\sqrt{3}}{2}$ $\sin {{60}^{\circ }}=\frac{\sqrt{3}}{2}$ $\cos {{60}^{\circ }}=\frac{1}{2}$