## Precalculus (6th Edition) Blitzer

The required solution is $\frac{\sqrt{3}}{2},\frac{1}{2},\frac{1}{2},\frac{\sqrt{3}}{2},0,1$
By putting the values according to the trigonometric functions, we get: $\cos {{30}^{\circ }}=\frac{\sqrt{3}}{2}$ $\sin {{30}^{\circ }}=\frac{1}{2}$ $\cos {{60}^{\circ }}=\frac{1}{2}$ $\sin {{60}^{^{\circ }}}=\frac{\sqrt{3}}{2}$ $\cos {{90}^{\circ }}=0$ $\sin {{90}^{\circ }}=1$