## Precalculus (6th Edition) Blitzer

The exact value of the provided expression is $-\frac{\sqrt{3}}{2}$.
The reference angle for $\theta =240{}^\circ$ is: \begin{align} & \theta '=240{}^\circ -180{}^\circ \\ & =60{}^\circ \end{align} Now, the function value for the reference angle is: $\sin 60{}^\circ =\frac{\sqrt{3}}{2}$ So, we see that the angle $240{}^\circ$ lies in quadrant III, so the sine function is negative: \begin{align} & \sin 240{}^\circ =-\sin 60{}^\circ \\ & =-\frac{\sqrt{3}}{2} \end{align}