Answer
The negative square root should be selected.
$$\sin195^\circ=-\sqrt{\frac{1-\cos390^\circ}{2}}$$
Work Step by Step
$$\sin195^\circ=\pm\sqrt{\frac{1-\cos390^\circ}{2}}$$
Whether the positive or negative square root should be selected depends the sign of $\sin195^\circ$.
$195^\circ$ lies in quadrant III. In quadrant III, $\sin\theta\lt0$. Therefore, $\sin195^\circ\lt0$.
Thus, the negative square root should be selected. In other words,
$$\sin195^\circ=-\sqrt{\frac{1-\cos390^\circ}{2}}$$
