Answer
$\sin{\theta} =-\dfrac{2\sqrt{5}}{5}$
$\cos{\theta} = -\dfrac{\sqrt{5}}{5}$
Work Step by Step
$y = 2x \\ \therefore \dfrac{y}{x} = 2 \\ \because \tan{\theta} = \dfrac{y}{x} \\ \therefore \tan{\theta} = 2$
$\because$ The terminal side of $\theta$ lies in $QIII%$, then $x$ and $y$ are both negative.
$\therefore $ Let $y=-2$ and $x =-1$
$r = \sqrt{x^2+y^2} = \sqrt{(-1)^2+(-2)^2} = \sqrt{5}$
$\sin{\theta} = \dfrac{y}{r} = \boxed{-\dfrac{2\sqrt{5}}{5}}$
$\cos{\theta} = \dfrac{x}{r} = \boxed{-\dfrac{\sqrt{5}}{5}}$
