Rozwiązanie

$\mathrm{tg}\alpha =-3$
$\frac{\sin \alpha }{\cos \alpha }=-3|\cdot \cos \alpha$
$\sin \alpha =-3\cos \alpha$

${\sin }^2\alpha +{\cos }^2\alpha =1$
${\left(-3\cos \alpha \right)}^2+{\cos }^2\alpha =1$
$9{\cos }^2\alpha +{\cos }^2\alpha =1$
$10{\cos }^2\alpha =1|:10$
${\cos }^2\alpha =\frac{1}{10}$   i   $\cos \alpha <0$, bo $\alpha$ jest kątem rozwartym
$\cos \alpha =-\frac{1}{\sqrt{10}}=\underline{-\frac{\sqrt{10}}{10}}$