Answer
$ -\dfrac{\sqrt{{2+\sqrt 3}}}{2}$
Work Step by Step
The half-angle Identity for cosine can be expressed as:
$\cos {(\dfrac{\theta}{2})}=\pm\sqrt{\dfrac{1+\cos{\theta}}{2}}$
$\cos (\dfrac{330^{\circ}}{2})=\cos 165^{\circ}$
The angle $165^{\circ}$ lies in the second quadrant, so we take the negative value of cosine.
Therefore,
$\cos{(165^{\circ})}=- \sqrt{\dfrac{1+\cos{(330^{\circ})}}{2}}\\=-\sqrt{\dfrac{1+\dfrac{\sqrt 3}{2}}{2}}\\=- \sqrt{\dfrac{2+\sqrt 3}{4}}\\ = -\dfrac{\sqrt{{2+\sqrt 3}}}{2}$
