## Precalculus (6th Edition) Blitzer

$-\frac{63}{65}$
Step 1. Given $sin\alpha=\frac{4}{5}, \frac{\pi}{2}\lt\alpha\lt\pi$, we can identify $\alpha$ to be in quadrant II, which means $cos\alpha=-\sqrt {1-sin^2\alpha}=-\frac{3}{5}$ Step 2. Given $cos\beta=\frac{5}{13}, 0\lt\beta\lt \frac{\pi}{2}$, we can identify $\beta$ to be in quadrant I, which means $sin\beta=\sqrt {1-cos^2\beta}=\frac{12}{13}$ Step 3. Use the the addition formula, we have $cos(\alpha+\beta)=cos\alpha cos\beta-sin\alpha sin\beta=(-\frac{3}{5})(\frac{5}{13})-(\frac{4}{5})(\frac{12}{13})=-\frac{63}{65}$