Answer
a. $\frac{4\sqrt {17}}{17}$
b. $-\frac{\sqrt {17}}{17}$
c. $ -4$
Work Step by Step
Given $tan\alpha=\frac{8}{15}$ and $180^\circ \lt \alpha \lt 270^\circ$, form a right triangle with sides $8,15,17$. We have $cos\alpha=-\frac{15}{17}$ and $90^\circ \lt \frac{\alpha}{2} \lt 135^\circ$
a. Use the Half-Angle Formula
$sin\frac{\alpha}{2}=\sqrt {\frac{1-cos\alpha}{2}}=\sqrt {\frac{1-(-\frac{15}{17})}{2}}=\frac{4\sqrt {17}}{17}$
b. Use the Half-Angle Formula
$cos\frac{\alpha}{2}=-\sqrt {\frac{1+cos\alpha}{2}}=\sqrt {\frac{1+(-\frac{15}{17})}{2}}=-\frac{\sqrt {17}}{17}$
c. $tan\frac{\alpha}{2}=\frac{sin\frac{\alpha}{2}}{cos\frac{\alpha}{2}}=-4$
(or use the Half-Angle Formula to get the same result.)
