How can I convert these parametric equations into a Cartesian equation (I think that is what its called). Wolfram alpha doesn't seem like it can handle this one.
$$
x=\cos{t}+\cos{-6t}\\
y=\sin{t}+\sin{-6t}
$$
Is there a name for this operation?
Converting parametric equation to Cartesian
algebraic-geometryparametric
Best Answer
I think you're looking for an implicit equation for this parametric plane curve. The process of turning a system of parametric equations into a system of implicit equations is called implicitization. See $\S3.3$ of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms for more on this.
Let $u = \cos(t)$ and $v = \sin(t)$. Using de Moivre's formula, we have \begin{align*} \cos(6t) &= T_6(\cos(t)) = 32u^6 - 48u^4 + 18u^2 - 1\\ \sin(6t) &= \sin(t) \, U_6(\cos(t)) = v(32u^5 - 32u^3 + 6u) \end{align*} where $T_6$ and $U_6$ are the Chebyshev polynomials of the first- and second kind, respectively. (Depending on your indexing, the second polynomial might be $U_5$.)
We form the ideal $$ (u^2 + v^2 - 1, x - (u + 32u^6 - 48u^4 + 18u^2 - 1), y - (v - v(32u^5 - 32u^3 + 6u))) $$ in $\mathbb{Q}[u,v,x,y]$ and compute a Gröbner basis under the elimination ordering. (This can be done in SageMath.) The last generator in this basis is \begin{align*} f(x,y) &:= x^{12} + 6 x^{10} y^{2} - 5 x^{10} + 15 x^{8} y^{4} - 25 x^{8} y^{2} + 5 x^{8} - 2 x^{7} + 20 x^{6} y^{6} - 50 x^{6} y^{4}\\ & \qquad + 20 x^{6} y^{2} + 42 x^{5} y^{2} + 15 x^{4} y^{8} - 50 x^{4} y^{6} + 30 x^{4} y^{4} - 70 x^{3} y^{4} + 6 x^{2} y^{10}\\ & \qquad - 25 x^{2} y^{8} + 20 x^{2} y^{6} + 14 x y^{6} + y^{12} - 5 y^{10} + 5 y^{8} \end{align*} so the implicit equation of the curve is $f(x,y) = 0$. Here are the parametric plot and implicit plot for comparison.