## Precalculus (6th Edition) Blitzer

a. $-\frac{240}{289}$ b. $-\frac{161}{289}$ c. $\frac{240}{161}$
Given $sin\theta=\frac{15}{17}$ and $\theta$ in quadrant II, form a right triangle with sides $15,8,17$. We have $cos\theta=-\frac{8}{17}$ and $tan\theta=-\frac{15}{8}$ a. $sin2\theta=2sin\theta cos\theta=2(\frac{15}{17})(-\frac{8}{17})=-\frac{240}{289}$ b. $cos2\theta=1-2sin^2\theta=1-2(\frac{15}{17})^2=-\frac{161}{289}$ c. $tan2\theta=\frac{sin2\theta}{cos2\theta}=\frac{240}{161}$ (or use the Double Angle Formula to get the same result.)