Answer
$-\dfrac{\sqrt3}{2}$
Work Step by Step
Convert the angle measure to degrees to obtain:
$=-\frac{8\pi}{3} \cdot \frac{180^o}{\pi} = -8(60^o)=-480^o$
Thus,
$\sin{(-\frac{8\pi}{3})} = \sin{(-480^o)}$
$-480^o$ is co-terminal with $-480^o+720^o=240^o$.
$240^o$ is in Quadrant III so its reference angle is $=240^o-180^o=60^o$.
Note that the sine function is negative in Quadrant III.
From Section 2.1 (page 50) , we learned that:
$\sin{60^o} = \dfrac{\sqrt3}{2}$
This means that:
$\sin{(-\frac{8\pi}{3})}
\\=\sin{(-480^o)}
\\=-\sin{60^o}
\\= -\dfrac{\sqrt3}{2}$
