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I know that a curve of the type $$\vec{\sigma}(t)=\cos(mt)\hat e_1+\cos(nt)\hat e_2$$ with $m,n\in\mathbb{Z}$ is algebraic.

My question is: what is the polynomial that define this curve?

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$\cos(nt)=T_n(\cos t)$, where $T_n$ is the $n$-th Chebyshev polynomial.

So your curve can be written $x=T_m(u)$ and $y=T_n(u)$, where $u=\cos(t)$.

Since $T_n(x)=T_n(T_m(u))=T_{nm}(u)=T_{mn}(u)=T_m(T_n(u))=T_m(y)$, your curve is (part of) the algebraic curve given by $T_n(x)=T_m(y)$.

This is not the lowest possible degree. For instance, if $lcm(m,n)=am=bn$, then $T_b(x)=T_a(y)$ is an equation of lower degree if $gcd(m,n)>1$ because $a=n/d$, $b=m/d$, $d=gcd(m,n)$.

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