## Precalculus (6th Edition) Blitzer

a. $\frac{3\sqrt {13}}{13}$ b. $\frac{2\sqrt {13}}{13}$ c. $\frac{3}{2}$
Given $sec(\alpha)=-\frac{13}{5}$ and $\frac{\pi}{2} \lt \alpha \lt \pi$, form a right triangle with sides $12,5,13$. We have $cos\alpha=-\frac{5}{13}$ and $\frac{\pi}{4} \lt \frac{\alpha}{2} \lt \frac{\pi}{2}$ a. Use the Half-Angle Formula $sin\frac{\alpha}{2}=\sqrt {\frac{1-cos\alpha}{2}}=\sqrt {\frac{1-(-\frac{5}{13})}{2}}=\frac{3\sqrt {13}}{13}$ b. Use the Half-Angle Formula $cos\frac{\alpha}{2}=-\sqrt {\frac{1+cos\alpha}{2}}=\sqrt {\frac{1+(-\frac{5}{13})}{2}}=\frac{2\sqrt {13}}{13}$ c. $tan\frac{\alpha}{2}=\frac{sin\frac{\alpha}{2}}{cos\frac{\alpha}{2}}=\frac{3}{2}$ (or use the Half-Angle Formula to get the same result.)