## Precalculus (6th Edition) Blitzer

a. $-\frac{15+8\sqrt 3}{34}$ b. $-\frac{8-15\sqrt 3}{34}$ c. $\frac{489-289\sqrt {3}}{33}$
Given $cos\alpha=\frac{8}{17}$ with $\alpha$ in quadrant IV, build a right triangle with angle $\alpha$ and sides $15,8,17$; we have $sin\alpha=-\frac{15}{17}$ and $tan\alpha=-\frac{15}{8}$. Similarly, given $sin\beta=-\frac{1}{2}$ with $\beta$ in quadrant III, build a right triangle with angle $\beta$ and sides $1,\sqrt {3},2$; we have $cos\beta=-\frac{\sqrt {3}}{2}$ and $tan\beta=\frac{\sqrt {3}}{3}$. Using the Addition Formulas, we have: a. $cos(\alpha+\beta)=cos(\alpha) cos(\beta)-sin(\alpha) sin(\beta)=(\frac{8}{17})(-\frac{\sqrt 3}{2})-(-\frac{15}{17})(-\frac{1}{2})=-\frac{15+8\sqrt 3}{34}$ b. $sin(\alpha+\beta)=sin(\alpha) cos(\beta)+cos(\alpha) sin(\beta)=(-\frac{15}{17})(-\frac{\sqrt {3}}{2 })+(\frac{8}{17})(-\frac{1}{2})=-\frac{8-15\sqrt 3}{34}$ c. $tan( \alpha+\beta)= \frac{sin(\alpha+\beta)}{cos(\alpha+\beta)}=\frac{8-15\sqrt 3}{15+8\sqrt 3}=\frac{489-289\sqrt {3}}{33}$