Answer
$-\dfrac{\sqrt{{2 -\sqrt 3}}}{2}$
Work Step by Step
The half-angle Identity for cosine can be expressed as:
$\sin {(\dfrac{\theta}{2})}=\pm\sqrt{\dfrac{1-\cos{\theta}}{2}}$
$\sin (\dfrac{390^{\circ}}{2})=\sin 195^{\circ}$
The angle $195^{\circ}$ lies in the third quadrant, so we take the negative value of sine.
Therefore, $\sin{(190^{\circ})}=- \sqrt{\dfrac{1 -\sin{(390^{\circ})}}{2}}\\=-\sqrt{\dfrac{1-\dfrac{\sqrt 3}{2}}{2}}\\=- \sqrt{\dfrac{2 -\sqrt 3}{4}}\\ = -\dfrac{\sqrt{{2 -\sqrt 3}}}{2}$
