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