I want to know how to prove that the graph of $r=\sin(\frac{\theta}{2})$ is symmetric about the $x$-axis (polar axis). As I understand it, if a polar graph is symmetrical about the $x$-axis, $(r,\theta)$ and $(r,-\theta)$ will give the same equation.
Since $r=\sin(\frac{-\theta}{2})=-\sin(\frac{\theta}{2})$, this is not the same as $r=\sin(\frac{\theta}{2})$.
On
This is an example of a problem where it is important to know how the polar curve is "traced out" in order to discuss questions about its polar function. For $ \ r \ = \ \sin \left(\frac{\theta}{2} \right) \ \ , \ $ we obtain a sort of "double cardioid" for which the "left half" is traced for the interval $ \ 0 \ < \ \theta \ < \ 2 \pi \ $ and the "right half" over $ \ 2 \pi \ < \ \theta \ < \ 4 \pi \ \ ; \ $ the curve begins at the origin for $ \ \theta \ = \ 0 \ \ , \ $ crosses through it at $ \ \theta \ = \ 2 \pi \ \ , \ $ and returns at $ \ \theta \ = \ 4 \pi \ \ $ to complete the cycle.
So the choice of $ \ -\theta \ $ to find the point which is the $ \ x-$axis "reflection" of the point $ \ (r \ = \ f(\theta) \ , \ \theta) \ $ will not work here. Since the $ \ x-$intercepts of this polar curve are $ \ (1 \ , \ \pi) \ $ and $ \ (-1 \ , \ 3 \pi) \ \ $ (more about "negative radii" shortly), the point $ \ \left( \ -\sin \left(\frac{\theta}{2} \right) \ , \ -\theta \ \right) \ $ is not the "vertical reflection" of $ \ \left( \ \sin \left(\frac{\theta}{2} \right) \ , \ \theta \ \right) \ \ , \ $ but is instead at a distance $ \ +\sin \left(\frac{\theta}{2} \right) \ $ from the origin in the direction $ \ \pi - \theta \ $ exactly opposite to $ \ -\theta \ \ , \ $ as follows from the way points at a "negative radius" from the origin are defined. (As examples, note that $ \ (-1 \ , \ -\pi = 3 \pi) \ $ or $ \ \left( -\frac12 \ , \ -\frac{\pi}{3} = \frac{5\pi}{3} \ \right) \ $ end up as "horizontal reflections" of $ \ (1 \ , \ \pi ) \ $ and $ \ \left( \frac12 \ , \ \frac{\pi}{3} \ \right) \ \ , \ $ respectively.)
So how are we to show that this polar curve is symmetrical about the $ \ x-$axis? Since the $ \ x-$axis passes through the curve at $ \ \theta \ = \ \pi \ $ and $ \ \theta \ = \ 3 \pi \ \ , \ $ the "vertically reflected" point to $ \ \pi - \theta \ $ lies at the angle $ \pi + \theta \ $ (and similarly for the points at $ \ 3 \pi - \theta \ $ and $ \ 3 \pi + \theta \ \ ) \ . \ $ We can check that $$ r \ \ = \ \ \sin \left(\frac{\pi \ - \ \theta}{2} \right) \ \ = \ \ \sin \left(\frac{\pi}{2} \right) \cos \left(\frac{\theta}{2} \right) \ - \ \cos \left(\frac{\pi}{2} \right) \sin \left(\frac{\theta}{2} \right) \ \ = \ \ \cos \left(\frac{\theta}{2} \right) $$ $$ = \ \ \sin \left(\frac{\pi \ + \ \theta}{2} \right) \ \ = \ \ \sin \left(\frac{\pi}{2} \right) \cos \left(\frac{\theta}{2} \right) \ + \ \cos \left(\frac{\pi}{2} \right) \sin \left(\frac{\theta}{2} \right) \ \ = \ \ \cos \left(\frac{\theta}{2} \right) \ \ ; $$
and likewise for $ \ 3 \pi - \theta \ $ and $ \ 3 \pi + \theta \ \ $ (with $ \ r \ $ being equal to $ \ -\cos \left(\frac{\theta}{2} \right) \ \ . \ ) $
You can use $\theta\mapsto-\theta$ to show it's symmetric about the $y$-axis.
And you can use $\theta\mapsto\theta+2\pi$ to show it's symmetric about the origin.
Even if you're restricting to $\theta\in[0,2\pi)$ (where it's no longer symmetric about the $y$-axis) and each substitution above is individually invalid, when you put them together...