Can all Lissajous curves be described by an implicit equation? I was trying to find the implicit form of the parametric curve $L(x(t),y(t))$ where $x(t)=5\cos\left(3 \pi t +\frac{\pi}{4}\right)$ and $y(t)=3cos\left(6 \pi t -\frac{\pi}{4}\right)$. At the end I wasted a lot of time and paper because I couldn't figure out the implicit expression (maybe I just didn't manipulate it properly). On the other hand, I saw a paper where these curves where described using Chebyshev's polynomials. I just want to know if the implicit form can be found without recurring to these polynomials.
 A: These curves can always be given by an implicit relation. If you have
a curve given by
$$x=a\cos(mt+c)$$
$$y=b\cos(nt+d)$$
with $m$ and $n$ positive integers, then you can use addition formulae
for sine and cosine to rewrite as
$$x=f(\cos t,\sin t)$$
$$y=g(\cos t,\sin t)$$
for some polynomials $f$ and $g$. If we let $u=\tan (t/2)$ we can use the half-angle
formulae $\cos t=(1-u^2)/(1+u^2)$ and $\sin t=2u/(1+u^2)$ to rewrite these
in the form
$$(1+u^2)^Mx=F(u)$$
$$(1+u^2)^Ny=G(u)$$
for polynomials $F$ and $G$. Now eliminate $u$ from these equations (say by resultants
or Groebner bases) to get a polynomial equation
$$H(x,y)=0.$$
I don't recommend you do this by hand....
A: I'd break it down like:
$$
\begin{align}
\frac{x}{5}
&=\cos(3\pi t + \pi/4)\\
&=\cos(3\pi t)\cos(\pi/4)-\sin(3\pi t)\sin(\pi/4)\\
&=kc-ks&k=1/\sqrt{2},c=\cos(3\pi t),s=\sin(3\pi t)
\end{align}
$$
and
$$
\begin{align}
\frac{y}{3}
&=\cos(6\pi t - \pi/4)\\
&=\cos(6\pi t)\cos(\pi/4)+\sin(6\pi t)\sin(\pi/4)\\
&=\left(\cos^2(3\pi t)-\sin^2(3\pi t)\right)\cos(\pi/4)+2\sin(3\pi t)\cos(3\pi t)\sin(\pi/4)\\
&=kc^2-ks^2+2kcs
\end{align}
$$
So now given these two relations along with $c^2+s^2=1$, you'd like to eliminate $c$ and $s$. It seems helpful to square sides of the first relation so it can be quadratic in $c$ and $s$ like the other two. Normalize by $k^2$, and by $k$ in the second relation.
$$
\begin{align}
\left(\frac{x}{5k}\right)^2&= c^2  +s^2  - 2cs
&\frac{y}{3k}&=                 c^2  -s^2  + 2cs
&1&=                            c^2  +s^2
\end{align}
$$
This is now three equations with three things to eliminate ($c^2,s^2,cs$). We'd like to have more equations than things to eliminate. To that end, square the sides of each equation. Also take each pairwise product of two left sides, equal to the corresponding product of right sides. This makes six equations:
$$
\begin{align}
\left(\frac{x}{5k}\right)^4  &= c^4 - 4c^3s + 6c^2s^2 - 4cs^3 + s^4 \\
\left(\frac{y}{3k}\right)^2  &= c^4 + 4c^3s + 2c^2s^2 - 4cs^3 + s^4 \\
1                            &= c^4         + 2c^2s^2         + s^4 \\
\left(\frac{x}{5k}\right)^2\frac{y}{3k}
                             &= c^4         - 4c^2s^2 + 4cs^3 - s^4 \\
\left(\frac{x}{5k}\right)^2  &= c^4 - 2c^3s + 2c^2s^2 - 2cs^3 + s^4 \\
\frac{y}{3k}                 &= c^4 + 2c^3s           + 2cs^3 - s^4
\end{align}
$$
Now we can try to use linear algebra to find a relation on the six vectors from $\mathbb{R}^5$ suggested by the right sides. Row reduction leads to:
$$-2(\text{right}_1)-(\text{right}_2)-(\text{right}_3)-2(\text{right}_4)+4(\text{right}_5)+2(\text{right}_6)=0$$
Therefore
$$-2\left(\frac{x}{5k}\right)^4-\left(\frac{y}{3k}\right)^2-1-2\left(\frac{x}{5k}\right)^2\frac{y}{3k}+4\left(\frac{x}{5k}\right)^2+2\frac{y}{3k}=0$$
Here is that implicit plot and here is the original parametric plot, confirming a match.
