## Trigonometry (11th Edition) Clone

$cos(\frac{\theta}{2}) = \sqrt{\frac{1}{5}}$
If $90^{\circ} \lt \theta \lt 180^{\circ}$, then the angle $\theta$ is in quadrant II. Then $45^{\circ} \lt \frac{\theta}{2} \lt 90^{\circ}$, so $\frac{\theta}{2}$ is in quadrant I. $cos(\frac{\theta}{2}) = \sqrt{\frac{1+cos~\theta}{2}}$ $cos(\frac{\theta}{2}) = \sqrt{\frac{1+(-\frac{3}{5})}{2}}$ $cos(\frac{\theta}{2}) = \sqrt{\frac{(\frac{2}{5})}{2}}$ $cos(\frac{\theta}{2}) = \sqrt{\frac{1}{5}}$