## Precalculus (6th Edition)

$-\sqrt {\dfrac {3}{5}}$
$\tan \theta =-\dfrac {\sqrt {6}}{2}$ (negative) $\cos \theta > 0$ (positive) So the angle is located in quadrant 4 ( sine is negative in quadrant 4) $\sin \theta =\dfrac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}=\dfrac {-\dfrac {\sqrt {6}}{2}}{\sqrt {1+\left( -\dfrac {\sqrt {6}}{2}\right) ^{2}}}=-\dfrac {\sqrt {6}}{\sqrt {10}}=-\sqrt {\dfrac {3}{5}}$