Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\sin x = \frac{1}{4}$,where $x$ is in quadrant $II$.

Given $x$ is in quadrant $II$,so $\frac{\pi}{2} < x < \pi$.
Dividing by $2$,we get $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.
Thus,$\frac{x}{2}$ lies in quadrant $I$,so $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are all positive.
Given $\sin x = \frac{1}{4}$.
Since $x$ is in quadrant $II$,$\cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - (\frac{1}{4})^2} = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}$.
Using half-angle formulas:
$\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-\frac{\sqrt{15}}{4})}{2}} = \sqrt{\frac{4 + \sqrt{15}}{8}} = \frac{\sqrt{8 + 2\sqrt{15}}}{4} = \frac{\sqrt{(\sqrt{5} + \sqrt{3})^2}}{4} = \frac{\sqrt{5} + \sqrt{3}}{4}$.
$\cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + (-\frac{\sqrt{15}}{4})}{2}} = \sqrt{\frac{4 - \sqrt{15}}{8}} = \frac{\sqrt{8 - 2\sqrt{15}}}{4} = \frac{\sqrt{(\sqrt{5} - \sqrt{3})^2}}{4} = \frac{\sqrt{5} - \sqrt{3}}{4}$.
$\tan \frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{(\sqrt{5} + \sqrt{3})^2}{5 - 3} = \frac{5 + 3 + 2\sqrt{15}}{2} = \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$.