Answer
a. $ -\frac{120}{169}$
b. $ -\frac{119}{169}$
c. $ \frac{120}{119}$
Work Step by Step
Given $sin\theta=\frac{12}{13}$ and $\theta$ in quadrant II, form a right triangle with sides $12,5,13$. We have $cos\theta=-\frac{5}{13}$ and $tan\theta=-\frac{12}{5}$
a. $sin2\theta=2sin\theta cos\theta=2(\frac{12}{13})(-\frac{5}{13})=-\frac{120}{169}$
b. $cos2\theta=1-2sin^2\theta=1-2(\frac{12}{13})^2=-\frac{119}{169}$
c. $tan2\theta=\frac{sin2\theta}{cos2\theta}=\frac{120}{119}$
(or use the Double Angle Formula to get the same result.)
