Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ given that $\cos x = -\frac{1}{3}$ and $x$ is in quadrant $III$.

(A) Given $x$ is in quadrant $III$,we have $\pi < x < \frac{3\pi}{2}$.
Dividing by $2$,we get $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$.
In this interval (quadrant $II$),$\sin \frac{x}{2}$ is positive,while $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are negative.
Using the identity $\cos x = 1 - 2\sin^2 \frac{x}{2}$,we have $\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} = \frac{1 - (-1/3)}{2} = \frac{4/3}{2} = \frac{2}{3}$.
Since $\sin \frac{x}{2} > 0$,$\sin \frac{x}{2} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$.
Using the identity $\cos x = 2\cos^2 \frac{x}{2} - 1$,we have $\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2} = \frac{1 + (-1/3)}{2} = \frac{2/3}{2} = \frac{1}{3}$.
Since $\cos \frac{x}{2} < 0$,$\cos \frac{x}{2} = -\sqrt{\frac{1}{3}} = -\frac{\sqrt{3}}{3}$.
Finally,$\tan \frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \frac{\sqrt{2}/\sqrt{3}}{-1/\sqrt{3}} = -\sqrt{2}$.