If $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$,find the values of $\sin \frac{x}{2}$,$\cos \frac{x}{2}$,and $\tan \frac{x}{2}$.

Given $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$.
Since $x$ is in the third quadrant,$\cos x$ is negative.
Using $\sec^2 x = 1 + \tan^2 x = 1 + \frac{9}{16} = \frac{25}{16}$,we get $\cos^2 x = \frac{16}{25}$.
Since $\cos x < 0$,$\cos x = -\frac{4}{5}$.
Given $\pi < x < \frac{3\pi}{2}$,dividing by $2$ gives $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$.
In this interval (second quadrant),$\sin \frac{x}{2} > 0$ and $\cos \frac{x}{2} < 0$.
Using half-angle formulas:
$2 \sin^2 \frac{x}{2} = 1 - \cos x = 1 - (-\frac{4}{5}) = \frac{9}{5} \implies \sin^2 \frac{x}{2} = \frac{9}{10} \implies \sin \frac{x}{2} = \frac{3}{\sqrt{10}}$.
$2 \cos^2 \frac{x}{2} = 1 + \cos x = 1 + (-\frac{4}{5}) = \frac{1}{5} \implies \cos^2 \frac{x}{2} = \frac{1}{10} \implies \cos \frac{x}{2} = -\frac{1}{\sqrt{10}}$.
Finally,$\tan \frac{x}{2} = \frac{\sin (x/2)}{\cos (x/2)} = \frac{3/\sqrt{10}}{-1/\sqrt{10}} = -3$.