## Precalculus (6th Edition) Blitzer

$\frac{3\sqrt {13}}{13}$
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 $cos\alpha=-\sqrt {1-sin^2\alpha}=-\frac{3}{5}$ and $tan\alpha=-\frac{4}{3}$ 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 $sin\beta=\sqrt {1-cos^2\beta}=\frac{12}{13}$ and $tan\beta=\frac{12}{5}$ Step 3. As $0\lt\frac{\beta}{2}\lt \frac{\pi}{4}$, using the half angle formula, we have $cos\frac{\beta}{2}=\sqrt {\frac{1+cos\beta}{2}}=\sqrt {\frac{1+(\frac{5}{13})}{2}}=\frac{3\sqrt {13}}{13}$