## Precalculus (6th Edition) Blitzer

a. $-\frac{4}{5}$ b. $\frac{5}{13}$ c. $\frac{33}{65}$ d. $-\frac{56}{65}$
a. Knowing $\alpha$ is in quadrant II, we have $cos\alpha\lt0$ and $cos\alpha=-\sqrt {1-sin^2\alpha}=-\frac{4}{5}$ b. Knowing $\beta$ is in quadrant II, we have $sin\beta\gt0$ and $sin\beta=-\sqrt {1-cos^2\beta}=\frac{5}{13}$ c. Using the addition formula, we have $cos(\alpha+\beta)=(-\frac{4}{5})(-\frac{12}{13})-(\frac{3}{5})(\frac{5}{13})=\frac{33}{65}$ d. Using the addition formula, we have $sin(\alpha+\beta)=(\frac{3}{5})(-\frac{12}{13})+(-\frac{4}{5})(\frac{5}{13})=-\frac{56}{65}$