## Precalculus (6th Edition)

$=-\dfrac {2\sqrt {5}}{5}$
Cosine is even function so $\cos \left( -\theta \right) =\dfrac {\sqrt {5}}{5}\Rightarrow \cos \theta =\dfrac {\sqrt {5}}{5}$ We see that $\cos \theta$ is positive and $\tan \theta$ is negative so angle is located in quadrant 4 and sine function is negative in quadrant 4 $\Rightarrow \sin \theta =-\sqrt {1-\cos ^{2}\theta }=-\sqrt {1-\left( \dfrac {\sqrt {5}}{5}\right) ^{2}}=-\dfrac {2\sqrt {5}}{5}$