Answer
a. $ \frac{720}{1681}$
b. $ \frac{1519}{1681}$
c. $ \frac{720}{1519}$
Work Step by Step
Given $sin\theta=-\frac{9}{41}$ and $\theta$ in quadrant III, form a right triangle with sides $9,40,41$. We have $cos\theta=-\frac{40}{41}$ and $tan\theta=\frac{9}{40}$
a. $sin2\theta=2sin\theta cos\theta=2(-\frac{9}{41})(-\frac{40}{41})=\frac{720}{1681}$
b. $cos2\theta=1-2sin^2\theta=1-2(-\frac{9}{41})^2=\frac{1519}{1681}$
c. $tan2\theta=\frac{sin2\theta}{cos2\theta}=\frac{720}{1519}$
(or use the Double Angle Formula to get the same result.)
