Answer
$\sin$ 1500$^{\circ}$ = $\frac{\sqrt3}{2}$
Work Step by Step
$\sin$ 1500$^{\circ}$
First, lets find the coterminal angle.
1500$^{\circ}$ - 4(360$^{\circ}$) = 60$^{\circ}$
Next, we find the reference angle.
$\theta$$^{1}$ = 60$^{\circ}$
Since the coterminal angle is in Quadrant I, $\sin$ is positive.
Therefore:
$\sin$ 60$^{\circ}$ = $\frac{\sqrt3}{2}$
