Algebra 2 (1st Edition)

$\dfrac{4 \sqrt 5}{9}$ and $\dfrac{1}{9}$ and $4 \sqrt 5$
$\cos a=-\sqrt {1-\sin^2 a}=-\sqrt {1-(-2/3)^2}=-\sqrt {\dfrac{5}{9}}=-\dfrac{\sqrt 5}{3}$ $\sin 2a=2 \sin a \cos a = 2 (-\dfrac{2}{3}) (-\dfrac{\sqrt 5}{3})=\dfrac{4 \sqrt 5}{9}$ $\cos 2a =\cos^2 a-\sin^2 a=(-\dfrac{\sqrt 5}{3})^2 - (-\dfrac{2}{3})^2=\dfrac{5}{9}-\dfrac{4}{9}=\dfrac{1}{9}$ $\tan 2a =\dfrac{\sin 2a}{\cos 2a}=\dfrac{4\sqrt 5/9}{1/9}=4 \sqrt 5$