Answer
a. $\frac{2\sqrt 5}{5}$
b. $-\frac{\sqrt 5}{5}$
c. $ -2$
Work Step by Step
Given $tan\alpha=\frac{4}{3}$ and $180^\circ \lt \alpha \lt 270^\circ$, form a right triangle with sides $4,3,5$. We have $cos\alpha=-\frac{3}{5}$ and $90^\circ \lt \frac{\alpha}{2} \lt 135^\circ$
a. Use the Half-Angle Formula
$sin\frac{\alpha}{2}=\sqrt {\frac{1-cos\alpha}{2}}=\sqrt {\frac{1-(-\frac{3}{5})}{2}}=\frac{2\sqrt 5}{5}$
b. Use the Half-Angle Formula
$cos\frac{\alpha}{2}=-\sqrt {\frac{1+cos\alpha}{2}}=\sqrt {\frac{1+(-\frac{3}{5})}{2}}=-\frac{\sqrt 5}{5}$
c. $tan\frac{\alpha}{2}=\frac{sin\frac{\alpha}{2}}{cos\frac{\alpha}{2}}=-2$
(or use the Half-Angle Formula to get the same result.)
