## Precalculus (6th Edition) Blitzer

The exact value of the provided expression is $-\frac{\sqrt{3}}{2}$.
We know that $\theta =\frac{22\pi }{3}$ lies in quadrant III, so the reference angle for $\theta =\frac{22\pi }{3}$ is: \begin{align} & \theta '=\frac{22\pi }{3}-\left( 6\pi +\pi \right) \\ & =\frac{22\pi -21\pi }{3} \\ & =\frac{\pi }{3} \end{align} Now, the function value for the reference angle is: $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$ Thus, we see that the angle $\theta =\frac{22\pi }{3}$ lies in quadrant III, so the sine function is negative: \begin{align} & \sin \frac{22\pi }{3}=-\sin \frac{\pi }{3} \\ & =-\frac{\sqrt{3}}{2} \end{align}