#### Answer

$-\dfrac {\sqrt {33}}{6}$

#### Work Step by Step

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