#### Answer

$-\displaystyle \frac{\sqrt{5}}{5}$

#### Work Step by Step

$\theta$ is in quadrant III, where both its sine and cosine are negative.
Furthermore,
$180^{o} < \theta < 270^{o}\qquad/\div 2$
$90^{o} < \displaystyle \frac{\theta}{2} < 135^{o}$,
so $\displaystyle \frac{\theta}{2}$ is in quadrant II, and $\displaystyle \cos\frac{\theta}{2}$ is negative.
Half-Angle Identity:
$\displaystyle \cos\frac{\theta}{2}=-\sqrt{\frac{1+\cos\theta}{2}}$
.. has the minus sign, we need $\cos\theta$ ...
Pythagorean identity, ($\cos x$ is negative):
$\cos\theta=-\sqrt{1-\sin^{2}\theta}=-\sqrt{1-\frac{16}{25}}$
$=-\displaystyle \sqrt{\frac{9}{25}}=-\frac{3}{5}=-0.6$
So,
$\displaystyle \cos\frac{\theta}{2}=-\sqrt{\frac{1+(-0.6)}{2}}$
$=-\displaystyle \sqrt{0.2}=-\sqrt{\frac{1}{5}}=-\frac{\sqrt{5}}{5}$