## Precalculus (6th Edition)

$-\displaystyle \frac{\sqrt{5}}{5}$
$\theta$ 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, $\displaystyle \cos\frac{\theta}{2}=-\sqrt{\frac{1+(-0.6)}{2}}$ $=-\displaystyle \sqrt{0.2}=-\sqrt{\frac{1}{5}}=-\frac{\sqrt{5}}{5}$