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If $(x^2+y^2)^3=4x^2y^2,$ then $r=\sin 2\theta$ for some $\theta$.

Using $r^2=x^2+y^2, x=r\cos\theta,y=r\sin\theta$, it's easy to get $r^2=\sin^22\theta$. But I don't know what to do next, since $r$ could be negative in $r=\sin2\theta.$

Actually the original problem is to show that the affine variety $V((x^2+y^2)^3-4x^2y^2)$ is contained in the four-leaved rose, whose polar equation is certainly $r=\sin 2\theta$. (Exercise 7(b), section 1.2, Ideals, Varieties and Algorithms, 3rd edition, David Cox etc.)

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    I assume it's the standard parametrization that $r = \sqrt{x^2 + y^2}$. In that case, if you end up with $r^2 = \sin^2{2\theta}$, there isn't any concern of $r$ being negative when you take the square root, since by definition $r$ is positive. Thus $r$ is the positive solution to $\sqrt{\sin^2{2\theta}}$, so indeed $r = \sin{\phi}$ for some $\phi$ (where that $\phi = 2\theta$). If this makes sense, I'd be happy to turn it into a full answer.2012-01-31
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    If we use one of the conventions (when $r$ is negative, reflect in the origin), the $r=\pm \sin 2\theta$ just traces out the figure twice.2012-01-31
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    @jamaicanworm, yes, $r$ must be positive, but not so $\sin2\theta$. The correct solution is $r=|\sin2\theta|$.2012-01-31
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    @Kou: If we decide that when $r<0$, the curve is not defined (which is one convention), then $r=\sin 2\theta$ has two leaves, not four.2012-01-31
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    @ArturoMagidin Well, I have put the original problem in my question. Did you see it? "to show that the affine variety..."2012-01-31
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    @AndréNicolas Could you explain that why $r=\sin 2\theta$ has two leaves when $r<0$?2012-01-31
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    @jamaicanworm The real original problem: Now argue carefully that $V((x^2+y^2)^3-4x^2y^2)$ is contained in the four-leaved rose. This is trickier than it seems since r can be negative in $r=\sin2\theta$.2012-01-31
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    @Kou: To repeat, there are two common conventions: (i) if $f(\theta)<0$, then the curve $r=f(\theta)$ is not defined at $\theta$ and (ii) if $f(\theta)<0$, find the point $(|f(\theta)|,\theta)$ and reflect it across the origin, or equivalently rotate $180$ degrees. Let $f(\theta)=\sin 2\theta$. Under convention (ii), we get $4$ leaves. Under convention (i), as $\theta$ travels $0$ to $90$, we get one leaf. In $(90,180)$, not defined, since there $\sin 2\theta$ is negative. In $[180, 270]$, new leaf. In $(270,360)$, not defined.2012-01-31
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    @AndréNicolas Thanks! I get it!2012-01-31
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    @Kou: So to sum up, unless you use convention (ii), $r=\sin 2\theta$ is not the four-leaved rose. If you want to use convention (i), then the response of Gerry Myerson is the right one. By the way, in lots of places you will see $r=\sin 2\theta$ called the four-leaf rose. That's because convention (ii) is quite commonly used.2012-01-31
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    @AndréNicolas But how to get $r=\sin 2\theta$ from that equation including $x$ and $y$, if we use convention (ii)? What is the trick here?2012-02-03
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    @Kou: Sorry for delay, I was away. Make the usual substitution. We get $r^6=4r^4\cos^2\theta\sin^2\theta=r^2\sin^2 2\theta$. Cancel the $r^4$. Technically, we can't do that, for we are missing the solution $r=0$. But $r=0$, $\theta=$ anything is one of the infinitely many addresses of the origin. The origin will turn out to be also on $r^2=\sin 2\theta$, so we will pick it up. If $r$ is allowed to be negative, then $r=\pm|\sin 2\theta|$. The curve $r=|\sin 2\theta|$ traces out the rose. The curve $r=-|\sin 2\theta|$ traces out the same rose. The end.2012-02-03
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    @Kou: the polar equation $r^2=\sin^2{2\theta}$ or $r=\pm\sin{2\theta}$ is, by symmetry ($\pi$-periodicity), geometrically equivalent to $r=|\sin{2\theta}|$. However, by an angle-doubling formula, this actually has period $\pi/2$, thus rising from and falling back to zero identically within each quadrant, to give the four-petaled shape. See my post below for more explanation. (I corrected my earlier mistake in the power or $r$, a remnant of which you see in my discussion of what would happen if we had $r=|\sin{2\theta}|^p$ for $p>0$ other than $1$, and in the blue curve for $p=1/3$ of my plot.)2012-02-08

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