## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\sqrt{\dfrac{(5-\sqrt 5)}{10}}$
We have: $\sin \theta=\dfrac{2}{\sqrt 5}$ and $\cos \theta=-\dfrac{1}{\sqrt 5}$ The half-angle Identity for cosine can be expressed as: $\cos {(\dfrac{\theta}{2})}=\pm \sqrt{\dfrac{1+\cos (\theta)}{2}}$ Because the angle $\dfrac{\theta}{2}$ lies in the first quadrant, we take the positive sign for cosine. $\cos {(\dfrac{\theta}{2})} = \sqrt{\dfrac{1+\cos (\dfrac{-1}{\sqrt 5})}{2}}=\sqrt{\dfrac{\sqrt 5 - 1}{2 \sqrt 5 }} = \sqrt{\dfrac{(5-\sqrt 5)}{10}}$