Answer
$\cos 2A=\frac{3}{5}$
$\cos \frac{A}{2}=-\sqrt {\frac{5-2\sqrt {5}}{10}}$
Work Step by Step
$\cos A=\pm\sqrt{1-\sin^{2}x}=\pm\sqrt {1-(-\frac{\sqrt 5}{5})^{2}}$
$=\pm\sqrt {1-\frac{5}{25}}=\pm\sqrt {\frac{4}{5}}=\pm\frac{2\sqrt 5}{5}$
Because $A$ is in the third quadrant and the cosine function is negative in that quadrant,
$\cos A=-\frac{2\sqrt 5}{5}$
$\cos 2A=\cos^{2}A-\sin^{2}A$
$=(-\frac{2\sqrt 5}{5})^{2}-(-\frac{\sqrt 5}{5})^{2}=\frac{3}{5}$
$\cos \frac{A}{2}=\pm\sqrt {\frac{1+\cos A}{2}}=\pm\sqrt {\frac{1+(-\frac{2\sqrt 5}{5})}{2}}$
$=\pm\sqrt {\frac{5-2\sqrt 5}{10}}$
As $180^{\circ}\leq A\leq270^{\circ}$,
$\frac{180^{\circ}}{2}\leq\frac{A}{2}\leq\frac{270^{\circ}}{2}$
Or $90^{\circ}\leq\frac{A}{2}\leq135^{\circ}$
$\frac{A}{2}$ terminates in the second quadrant. In the second quadrant, the cosine function is negative. So,
$\cos \frac{A}{2}=-\sqrt {\frac{5-2\sqrt 5}{10}}$
