Algebra 2 (1st Edition)

$-\dfrac{\sqrt 3}{2}$
Here, we have $\theta=-840^{\circ}$ Here, $\theta'=-840^{\circ}+(3) 360^{\circ}=240^{\circ}$ and $\theta''=\theta'-180^{\circ}=240^{\circ}-180^{\circ}=60^{\circ}$ The function $\sin$ is negative in Quadrant $II$. Now, $\sin \theta=\sin \theta'= -\sin \theta''=-\sin 60^{\circ}=-\dfrac{\sqrt 3}{2}$