## Precalculus (6th Edition)

$-\dfrac {\sqrt {33}}{6}$
$\cos \left( -\theta \right) =\cos \theta =\dfrac {\sqrt {3}}{6}$ (cosine is positive) $\cot \theta < 0$ ($\cot$ is negative) So angle is located in quadrant 4 and sine is negative in quadrant 4 $\Rightarrow \sin \theta =-\sqrt {1-\cos ^{2}\theta }=-\sqrt {1-\left( \dfrac {\sqrt {3}}{6}\right) ^{2}}=-\dfrac {\sqrt {33}}{6}$