Answer
a. $ \frac{4\sqrt 5}{9}$
b. $ \frac{1}{9}$
c. $ 4\sqrt 5 $
Work Step by Step
Given $sin\theta=-\frac{2}{3}$ and $\theta$ in quadrant III, form a right triangle with sides $2, \sqrt 5, 3$. We have $cos\theta=-\frac{\sqrt 5}{3}$ and $tan\theta=\frac{2\sqrt 5}{5}$
a. $sin2\theta=2sin\theta cos\theta=2(-\frac{2}{3})(-\frac{\sqrt 5}{3})=\frac{4\sqrt 5}{9}$
b. $cos2\theta=1-2sin^2\theta=1-2(-\frac{2}{3})^2=\frac{1}{9}$
c. $tan2\theta=\frac{sin2\theta}{cos2\theta}=4\sqrt 5 $
(or use the Double Angle Formula to get the same result.)
