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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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    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...2012-02-02

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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!2012-02-02