## Precalculus (6th Edition) Blitzer

a.) $\frac{\sqrt 3}{2}$ b.) -$\frac{\sqrt 3}{2}$
a.)sin$\frac{2\pi}{3}$ corresponds to y value. sin$\frac{2\pi}{3}$ = $\frac{\sqrt 3}{2}$ b.)Sine is an odd function, so sin(-t) = -sin t. sin-$\frac{2\pi}{3}$ = - sin$\frac{2\pi}{3}$ = -$\frac{\sqrt 3}{2}$