Under what circumstances do symmetries in polar graphs hold?

Question

It’s common knowledge that when dealing with polar graphs if you can replace r with -r and θ with -θ then the graph should be symmetrical about θ=pi/2.

However, when drawing r=θ/2 it seems this doesn’t hold true.

Are there specific criteria that need to hold for the standard 3 symmetry tests to hold? Also under what circumstances are the symmetries reflections or rotations?

Thanks in advance.

Answer 1

The 3 symmetry tests are necessary and sufficient conditions (in fact, definitions) on any set "closed under taking negatives." That is, if $\theta$ is in the domain, then $-\theta$ should be too. For example, this includes sets of the form $(-a, a)$, and the whole real line. If, for some angle $\theta_0$ in the domain of your function, the opposite angle $-\theta_0$ is not in the domain, then the conditions of the symmetry test don't even make sense, since $r(-\theta)$ isn't defined.

The function $r = \theta/2$ is symmetric about the line $\theta = \pi / 2$ as long as whenever you plot $r(\theta)$ you also plot $r(-\theta)$. If you plot it on $[0, a]$ for some real number $a$, you will notice that this symmetry does not exist.

As for the second question, each symmetry test deals with a specific type of symmetry and I will restate them here.

Let $r = f(\theta)$ be a function $\mathbb{R} \to \mathbb{R}$.

1. If, for each $\theta$, $r(\theta) = r(-\theta)$, then $f$ is equal to its reflection about the line $\theta = 0$ ($x$-axis).
2. If, for each $\theta$, $r(\theta) = -r(\theta)$, then $f$ is equal to its rotation by $\pi$ radians about the origin.
3. If, for each $\theta$, $r(\theta) = -r(-\theta)$, then $f$ is equal to its reflection about the line $\theta = \pi/2$ ($y$-axis).