Question on identifying symmetries in a polar graph

Question

I've tried asking this question before but I'm still confused.

Let's say I have the polar equation $r^2=\sin\theta$

For the symmetry of x axis I am replacing $\theta$ with $-\theta$ and I get:

$r^2=\sin(-\theta)=-\sin(\theta)$

For the y axis I am replacing $r$ with $-r$ and $\theta$ with $-\theta$ and I get:

$(-r)^2=\sin(-\theta)$

$r^2=-\sin(\theta)$

For the pole I replace $r$ with $-r$ and I get:

$(-r)^2=\sin(\theta)$

$r^2=\sin(\theta)$

The pole is the only one I understand because I get back to $r^2=\sin(\theta)$

I understand that for the x-axis, if I replace $\theta$ with $\pi-\theta$ and $r$ with $-r$ I will find that it is symmetric. And to find symmetry for the y-axis, I need to replace $\theta$ with $\pi-\theta$.

My question is, do I need to test both methods each time I am trying to find symmetries?

In other words, to test for symmetry in the x-axis, do I need to replace $\theta$ with $-\theta$ and if nothing results from that then I should also try replacing $\theta$ with $\pi-\theta$ and $r$ with $-r$?

What does it mean that when I tried replacing $\theta$ with $-\theta$, I didn't find any symmetry?

Also, when I plot points for this graph, I don't get anything from $\pi$ to $2\pi$ since the graph of $\sin\theta$ is negative here but $r^2$ cannot be negative. So how do I even have something from $\pi$ to $2\pi$ on the actual graph? I just have to rely on symmetries to know this? If I simply plotted points, or created an $r^2-\theta$ graph, I don't think I'd have enough information to graph this properly.

Answer 1

A function $f(r,\theta)$ is symmetric to the ...

• $x$-axis if $f(r,\theta) = f(r, -\theta)$ or $f(-r, \pi - \theta) \, \to r^2 - \sin(\theta) = (-r)^2 - \sin(\pi - \theta)$
• $y$-axis if $f(r,\theta) = f(-r, -\theta)$ or $f(r,\pi - \theta) \; \to r^2 - \sin(\theta) = r^2 - \sin(\pi - \theta)$
• origin if $f(r, \theta) = f(-r, \theta)$ or $f(r, \pi + \theta) \quad \to r^2 - \sin(\theta) = (-r)^2 - \sin(\theta)$

To rule some of these properties out you have to check both cases. For any $\theta$ there are two valus of $r$ that satisfy your equation:

$\hspace{4cm}$