Polar to cartesian form of $r=\tan(2θ)$

Question

I have attempted converting $r=\tan(2θ)$ to cartesian coordinates:

$$r=\frac{2\sin(θ)\cos(θ)}{\cos^2(θ)-\sin^2(θ)}$$ $$r=\frac{2r\sin(θ)r\cos(θ)}{r^2\cos^2(θ)-r^2\sin^2(θ)}$$

$r^2 = x^2 + y^2\\ x = r \cos \theta\\ y = r \sin \theta$

$$r=\frac{2xy}{x^2-y^2} $$ $$(x^2+y^2)^{1/2}=\frac{2xy}{x^2-y^2} $$ $$(x^2+y^2)= \left(\frac{2xy}{x^2-y^2}\right)^2 $$ This doesn't graph properly on Wolfram Alpha, so I must have made a mistake.

When in doubt, use Desmos. It is very good at plotting implicit curves, and it shows that your derivation is correct. — Parcly Taxel, Sep 16 '16 at 16:09
@Jean-ClaudeArbaut Are you saying something different than the accepted answer? — User3910, Sep 16 '16 at 19:31
Yes, the accepted answer is actually wrong, and the polar plot $r=\tan(2\theta)$ is indeed, the full graph of $x^2+y^2=(2xy/(x^2-y^2))^2$, though it's not completely obvious. — Jean-Claude Arbaut, Sep 16 '16 at 19:50
By the way, here is how to ask WA: cartesian, polar. Remove the straight lines from the polar plot (they appear because of jumps at tangent discontinuities) — Jean-Claude Arbaut, Sep 16 '16 at 20:01
@Jean-ClaudeArbaut I think your explanation is completely obvious now that the function is graphed correctly. Apparently the mistake I made leading to this quesiton was improper plotting. — User3910, Sep 16 '16 at 20:11
use that $$x=r\cos(\phi),y=r\sin(\phi),r=\sqrt{x^2+y^2}$$ and i got the same as you. — Dr. Sonnhard Graubner, Sep 16 '16 at 16:35

score 1 · Accepted Answer · answered Sep 16 '16 at 20:45

Not an answer, but an illustration.

For $\theta$ from $-\pi/4$ to $\pi/4$, you get the blue branch. It starts on the left for negative $\theta$. Because of the $\pi/2$ periodicity of $r$, the other branches are found by rotating the first by $\pi/2$ repeatedly.

Polar to cartesian form of $r=\tan(2θ)$

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