## Precalculus (6th Edition)

$\sin \theta =-\dfrac {3}{\sqrt {10}}$
$\sin \theta =\pm \dfrac {1}{\sqrt {1+\cot ^{2}\theta }}=\pm \dfrac {1}{\sqrt {1+\left( -\dfrac {1}{3}\right) ^{2}}}=\pm \dfrac {3}{\sqrt {10}}$ In quadrant 4 sine function is negative so $\sin \theta =-\dfrac {3}{\sqrt {10}}$