I encountered this when trying to prove polar-axis symmetry for $\sin \frac{\theta}{2}$, and ended up with:
$r(\pi-\theta)=\sin\left(\frac\pi2-\frac\theta2\right)=\cos\frac\theta2$
Implying that $\cos \frac{\theta}{2} = -\sin \frac{\theta}{2}$, per the symmetry test $(r,\theta) \rightarrow (-r, \pi-\theta)$. I noticed that while in the rectangular system, $\cos \frac{x}{2}$ and $-\sin \frac{x}{2}$ are different, but in the polar system, they are graphically identical. Why is this so, and how do I proceed to prove the symmetry algebraically?
$\cos \frac{\theta}{2}$ and $-\sin \frac{\theta}{2}$ are not at all the same thing. However, when you graph $r=\cos \frac{\theta}{2}$ and $r=-\sin \frac{\theta}{2}$ you will obtain the same graph with a distinct difference that the points on the graph are traced out differently. More precise: A point of one graph "trails" a point on the other graph by $\pi$. The reason for this is that polar points of the graph do not have a unique $(r,\theta)$ representation as what you would have with Cartesian coordinates. If you have two Ti 84's and you would graph both polar curves at the same time, you can see what I mean. A very simple example to illustrate: $r=1$ and $r=-1$ both represent the unit circle, but the first one starts at $(1,0)$, and the second one at $(-1,0)$ and then they both "move" counter clockwise. Same graph, but obviously we can't say that $-1$ and $1$ are equal