## Precalculus (6th Edition) Blitzer

a. $\frac{\sqrt {58}(15+7\sqrt {11})}{348}$ b. $-\frac{\sqrt {58}(35-3\sqrt {11})}{348}$ c. $-\frac{378-145\sqrt {11}}{74}$
Given $sin\alpha=\frac{5}{6}$ with $\alpha$ in quadrant II, build a right triangle with angle $\alpha$ and sides $5,\sqrt {11},6$; we have $cos\alpha=-\frac{\sqrt {11}}{6}$ and $tan\alpha=-\frac{5\sqrt {11}}{11}$. Similarly, given $tan\beta=\frac{3}{7}$ with $\beta$ in quadrant III, build a right triangle with angle $\beta$ and sides $3,7,\sqrt {58}$; we have $sin\beta=-\frac{3\sqrt {58}}{58}$ and $cos\beta=-\frac{7\sqrt {58}}{58}$. Using the Addition Formulas, we have: a. $cos(\alpha+\beta)=cos(\alpha) cos(\beta)-sin(\alpha) sin(\beta)=(-\frac{\sqrt {11}}{6})(-\frac{7\sqrt {58}}{58})-(\frac{5}{6})(-\frac{3\sqrt {58}}{58})=\frac{\sqrt {58}(15+7\sqrt {11})}{348}$ b. $sin(\alpha+\beta)=sin(\alpha) cos(\beta)+cos(\alpha) sin(\beta)=(\frac{5}{6})(-\frac{7\sqrt {58}}{58})+(-\frac{\sqrt {11}}{6})(-\frac{3\sqrt {58}}{58})=-\frac{\sqrt {58}(35-3\sqrt {11})}{348}$ c. $tan( \alpha+\beta)= \frac{sin(\alpha+\beta)}{cos(\alpha+\beta)}=-\frac{35-3\sqrt {11}}{15+7\sqrt {11}}=-\frac{378-145\sqrt {11}}{74}$