In polar plots $r=f(\theta)$, should a negative radial coordinate always be reflected? Even if $f(\theta+\pi)\neq-f(\theta)$? If $f$ does not involve any trigonometric functions?
For instance, is this correct?
Or is this correct?
My Argument
I think the first (where negative radial coordinates are ignored) is correct.
Our function is $f(\theta)=1-2\sin\theta$. Consider the point $(-1,\pi/2)$, which satisfies $r=f(\theta)$. The reflection rule tells us to transform it to $(1, 3\pi/2)$ and plot it. However, $f$ does not have the reflection property: in general, $f(\theta+\pi)\neq-f(\theta)$. Naturally, the transformed point does not satisfy $r=f(\theta)$. Why, then, should we plot it?
In other words, I think we should plot reflected points if and only if $f(\theta+\pi)\neq-f(\theta)$. This also makes sense if we consider non-trigonometric functions, like $f(\theta)=\theta-\pi$. (They also do not exhibit a sign-flip when rotated by $\pi$.) Is this intuition wrong?
Other Answers
Most demonstrations I have seen conveniently use simple functions with which a strong case is made for reflection. So, I haven't been able to find a satisfactory answer to this.
This answer seems to imply that we always reflect (since the statement about reflecting is made without reference to any function). On the other hand, this answer makes a case for not reflecting. I agree, but not for the same reasons (the author says that reflecting would make that particular function non-bijective).
Here's another answer that suggests reflecting without mentioning any specific functions. And another and another and another (but the last two are justified, because the question wasn't asked in context of plots or functions).
What should the graph of $$r=-\lvert\theta\rvert$$ look like? Taking the approach of omitting negative $r$ -- nothing, a blank plot. Taking the approach of reflecting the points, we have a spiral that shows symmetry with $r=\theta$.$^*$ Of the two of these, I think preserving the intuition that negating a result produces some kind of reflection should win out.
Nor does excluding negative $r$ solve the problem you are trying to fix. Imagine a graph of $r=\theta$ from $\theta=0$ to $\theta=6\pi$. Reading from points on the curve alone, it looks as if the points $(2\pi, 0)$, $(4\pi, 0)$ and $(6\pi, 0)$ are solutions to $r=\theta$. They are not, of course. What is actually happening is that, holding $r$ constant, $(r, \theta+2\pi k)$ names the same point for any integer choice of $k$. Points in polar form just aren't unique in the way points on the Cartesian plane are, and a little bit more inference from the viewer is required to figure out which integer multiple of $\theta$ is intended to be associated to which $r$. And once you accept you have to do that inferential work, what additional work are you saving yourself from doing by insisting that $r$ stay positive?