Answer
a. $\frac{2\sqrt {6}}{6}$
b. $\frac{\sqrt {3}}{3}$
c. $ \sqrt 2$
Work Step by Step
Given $sec(\alpha)=-3$ and $\frac{\pi}{2} \lt \alpha \lt \pi$, form a right triangle with sides $2\sqrt 2,1,3$. We have $cos\alpha=-\frac{1}{3}$ and $\frac{\pi}{4} \lt \frac{\alpha}{2} \lt \frac{\pi}{2}$
a. Use the Half-Angle Formula
$sin\frac{\alpha}{2}=\sqrt {\frac{1-cos\alpha}{2}}=\sqrt {\frac{1-(-\frac{1}{3})}{2}}=\frac{2\sqrt {6}}{6}$
b. Use the Half-Angle Formula
$cos\frac{\alpha}{2}=-\sqrt {\frac{1+cos\alpha}{2}}=\sqrt {\frac{1+(-\frac{1}{3})}{2}}=\frac{\sqrt {3}}{3}$
c. $tan\frac{\alpha}{2}=\frac{sin\frac{\alpha}{2}}{cos\frac{\alpha}{2}}=\sqrt 2$
(or use the Half-Angle Formula to get the same result.)
