Answer
The exact value of the expression is $\frac{\sqrt{3}}{2}$.
Work Step by Step
$\sin \left( -\frac{17\pi }{3} \right)$ lies in quadrant I.
Add $6\pi $ to $-\frac{17\pi }{3}$ , to find a positive co-terminal angle less than $2\pi $. Consider $\alpha $ to be the positive co-terminal angle.
$\begin{align}
& \alpha =\left( -\frac{17\pi }{3} \right)+6\pi \\
& =\frac{-17\pi +18\pi }{3} \\
& =\frac{\pi }{3}
\end{align}$
The positive acute angle formed by the terminal side of $\theta $ and the x-axis is the reference angle ${\theta }'$.
The reference angle of $-\frac{17\pi }{3}$ is $\frac{\pi }{3}$.
The function value of the reference angle is
$\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$
The angle $-\frac{17\pi }{3}$ is in quadrant I and the sine function is positive in quadrant I.
Therefore,
$\sin \left( -\frac{17\pi }{3} \right)=\sin \frac{\pi }{3}$
Substitute $\frac{\sqrt{3}}{2}$ for $\sin \frac{\pi }{3}$.
$\sin \left( -\frac{17\pi }{3} \right)=\frac{\sqrt{3}}{2}$
The exact value of the expression is $\frac{\sqrt{3}}{2}$.
