## Precalculus (6th Edition) Blitzer

a. $-\frac{4+3\sqrt {15}}{20}$ b. $\frac{-3+4\sqrt {15}}{20}$ c. $\frac{-192+25\sqrt {15}}{119}$
Given $tan\alpha=\frac{3}{4}$ with $\alpha$ in quadrant III, build a right triangle with angle $\alpha$ and sides $3,4,5$; we have $sin\alpha=-\frac{3}{5}$ and $cos\alpha=-\frac{4}{5}$. Similarly, given $cos\beta=\frac{1}{4}$ with $\beta$ in quadrant IV, build a right triangle with angle $\beta$ and sides $\sqrt {15},1,4$; we have $sin\beta=-\frac{\sqrt {15}}{4}$ and $tan\beta=-\sqrt {15}$. Using the Addition Formulas, we have: a. $cos(\alpha+\beta)=cos(\alpha) cos(\beta)-sin(\alpha) sin(\beta)=(-\frac{4}{5})(\frac{1}{4})-(-\frac{3}{5})(-\frac{\sqrt {15}}{4})=-\frac{4+3\sqrt {15}}{20}$ b. $sin(\alpha+\beta)=sin(\alpha) cos(\beta)+cos(\alpha) sin(\beta)=(-\frac{3}{5})(\frac{1}{4})+(-\frac{4}{5})(-\frac{\sqrt {15}}{4})=\frac{-3+4\sqrt {15}}{20}$ c. $tan( \alpha+\beta)=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha) tan(\beta)}=\frac{(\frac{3}{4})+(-\sqrt {15})}{1-(\frac{3}{4})(-\sqrt {15})}=\frac{3-4\sqrt {15}}{4+3\sqrt {15}}=\frac{-192+25\sqrt {15}}{119}$