Self-intersections of a Lissajous curve Hei,
I want to find the self intersections of a Lissajous curve, for instance:
$$x(t)=\sin 2t$$
$$y(t)=\sin 3t$$
I have been trying for a couple of hours but I really don't get how to compute all the solutions. Basically I was trying to write the relation such that
$$\sin 2u=\sin 2v \text{ and }\sin 3u=\sin 3v.$$
Thanks for the support!
 A: Start with a rational parametrization, using the tangent half angle formula:
$$c=\cos u=\frac{1-a^2}{1+a^2}\qquad s=\sin u=\frac{2a}{1+a^2}$$
Use Wikipedia to find formulas for double and triple angle.
$$\sin 2u=2sc=\frac{(1-a^2)2a}{(1+a^2)^2}
\qquad
\sin3u=3s-4s^3=\frac{6a(1+a^2)^2-32a^3}{(1+a^2)^3}$$
Now assume a second parameter $b$ instead of $a$ and write your equations, cross-multiplying in the process:
$$(1-a^2)2a(1+b^2)^2=(1-b^2)2b(1+a^2)^2\\
(6a(1+a^2)^2-32a^3)(1+b^2)^3=(6b(1+b^2)^2-32b^3)(1+a^2)^3$$
Now you have a polynomial system of two equations in two variables. You can cancel the trivial solution $a=b$ from both of them (e.g. by writing them as LHS − RHS = $0$, then factorizing and removing the factor $a-b$). Then using a computer algebra system of your choice, or some elimination technique like resultants, you find 6 algebraic solutions ignoring duplicates. Note that the tangent half-angle substitution omits one point, namely $a=\infty$ corresponding to $t=\pi$. You need to check that case separately, and will indeed find one more solution there. Turning parameters back into points you get these 7 points of intersection:
\begin{align*}
x &= \tfrac{1}{2} \, \sqrt{3} & y &= 0 &
\{a,b\}&=\{
-\sqrt{3},
\sqrt{\tfrac{1}{3}}
\}\\
x &= -\tfrac{1}{2} \, \sqrt{3} & y &= 0 &
\{a,b\}&=\{
\sqrt{3},
-\sqrt{\tfrac{1}{3}}
\}\\
x &= \tfrac12 & y &= \tfrac12\sqrt2 &
\{a,b\}&=\{
-\sqrt{6} + \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} - 2,
\sqrt{6} - \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} - 2
\}\\
x &= \tfrac{1}{2} & y &= -\tfrac{1}{2} \, \sqrt{2} &
\{a,b\}&=\{
-\sqrt{6} - \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} - 2,
\sqrt{6} + \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} - 2
\}\\
x &= -\tfrac{1}{2} & y &= -\tfrac{1}{2} \, \sqrt{2} &
\{a,b\}&=\{
-\sqrt{6} + \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} + 2,
\sqrt{6} - \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} + 2
\}\\
x &= -\tfrac{1}{2} & y &= \tfrac{1}{2} \, \sqrt{2} &
\{a,b\}&=\{
-\sqrt{6} - \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} + 2,
\sqrt{6} + \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} + 2
\}\\
x &= 0 & y &= 0 & \{a,b\}&=\{0,\infty\}
\end{align*}
There are also complex solutions for $\{a,b\}=\{-i,+i\}$ but I assume these are not interesting to you.
