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