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The equation is:

$r^2=-4 \sin(2\theta)$

I first made a reference graph in cartesian coordinates using values $\displaystyle \frac{\pi}{4}$, $\displaystyle \frac{\pi}{2}$, $\displaystyle \frac{3 \pi}{4}$, $\displaystyle \pi$. Then from that I formed this:

enter image description here

Something seems off about that though. Should it be across the other axis instead?

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    If you plug in $\pi/4$, what does it mean for $r^2$ to be $-4$? (You treated it like $r=-4$, but even that is impossible.)2011-05-22

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You can let WolframAlpha plot this by rewriting it in Cartesian coordinates:

$r^2=-4\sin2\theta=-2\sin\theta\cos\theta\;,$ $r^4=-2\sin\theta r\cos\theta r=-2xy\;.$

Concerning your own plot: It seems it's not the axes you got mixed up, but the sine and cosine.

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    @joriki I guess continuing this discussion here would be a misuse of the comments. Let me just say that I partly agree with you, and in fact I myself teach my representation theory students how to compute things with MAGMA. Maybe we will get the opportunity some day to have a proper conversation on this topic offline.2011-05-22
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Give the comment from Alon some consideration, though I wasn't able to immediately eliminate your potential graph based on that idea alone. Let's think about the interval $\frac{\pi}{2}< \theta< \pi$. On that interval, $\sin2\theta$ goes from $0$ to $-1$ to $0$, so $r^2=-4\sin2\theta$ goes from $0$ to $4$ to $0$, so $r$ goes from $0$ to $\pm 2$ to $0$ again. No matter how I interpret your graph, it says that $r=\pm4$ at $\theta=\frac{\pi}{2}$ or $\theta=\frac{3\pi}{2}$ (or something along those lines).