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A need to show that the curve given in polar equation $\mathbb{r=a\sin(b\theta)}$ is an algebraic curve if $b=\frac{m}{n}$, $m,n\in \mathbb{N}^{*}$ and $(m,n)=1$. Also I am supposed to find the polynomial which it satisfies. If $b$ is an irrational number then the curve (I don't know the name in English) is not algebraic.

I would appreciate any help with this problem.

PS. I've studied Algebraic Curves, Commutative Algebra and Algebraic Geometry many years ago (7 years to be exact!). So, I am trying get back to business. If it is a silly question, please delete it.

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Indeed, only if $b\in \mathbb Q$ is $r=a\sin(b\theta)$ algebraic. There might be a slicker proof, but the trick is to start with the parametric equations $x=r(\theta)\cos\,\theta,\quad y=r(\theta)\sin\,\theta$, and taking $b=p/q$ with $b$ in lowest terms, express $x$ and $y$ entirely in terms of $\cos\dfrac{\theta}{q}$ and $\sin\dfrac{\theta}{q}$. Now, you can use Gröbner bases or some other elimination technique to derive your algebraic equation. – J. M. Feb 2 '12 at 0:47
@J.M. I completely forgot this stuff, so my question might be stupid. Anyhow, can't you get a simple expression involving only the Chebyshev polynomials ? – N. S. Feb 2 '12 at 1:30
Even if you did end up with expressing the parametric components entirely in terms of Chebyshev polynomials, I doubt that the implicit algebraic equation you can derive from those would admit something similarly compact... – J. M. Feb 2 '12 at 1:33
up vote 3 down vote accepted

Because of J.M.'s comment, I'll just show that if $b$ is irrational then the curve $r=\text a \sin b\theta$ is not algebraic.

Indeed the points corresponding to $\theta =2k\pi,\; k\in \mathbb N$ are all distinct points on the same ray, since $\sin (b\cdot 2k\pi)\neq (\sin b\cdot 2l\pi)$ for $k\neq l \in \mathbb N$
But an algebraic curve not containing a line can only cut that line in finitely many points.
[Because a non-zero polynomial has only finitely many zeros ]

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Ah, this was the clever approach I was hoping to see. +1! – J. M. Feb 2 '12 at 10:28

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