#### Answer

$\dfrac{1}{2}$

#### Work Step by Step

Convert the angle measure to degrees to obtain:
$=-\frac{7\pi}{6} \cdot \frac{180^o}{\pi} = -7(30^o)=-210^o$
Thus,
$\sin{(-\frac{7\pi}{6})} = \sin{(-210^o)}$
$-210^o$ is co-terminal with $-210^o+360^o=150^o$.
$150^o$ is in Quadrant II so its reference angle is $=180^o-150^o=30^o$.
Note that the sine function is positive in Quadrant II.