TITLE: How can the trefoil knot be expressed in polar coordinates?
QUESTION [9 upvotes]: From Wikipedia, the parametric equations for a trefoil knot are
\begin{align*}
x(t) &= \sin t + 2\sin 2t \\
y(t) &= \cos t - 2\cos 2t \\
z(t) &= -\sin 3t.
\end{align*}
I am only interested in the $x$ and $y$ dimensions, so $z(t)$ is ignored. When I plot it with Wolfram|Alpha, I get the expected general shape. However, when I try to convert it to polar coordinates, it (seemingly) just doesn't work.
\begin{align*}
r^2 &= x^2 + y^2 \\
&= (\sin t + 2\sin 2t)^2 + (\cos t + 2\cos 2t)^2 \\
&= (\sin^2 t + 4\sin t \sin 2t + 4\sin^2 2t) + (\cos^2 t - 4\cos t \cos 2t + 4\cos^2 2t) \\
&= 1+4 + 4(\sin t \sin 2t - \cos t \cos 2t) \\
&= 5-4\cos 3t
\end{align*}
Yet, when I try to plot $r = \sqrt{5-4\cos 3t}$, I get something completely different. What's the problem? Additionally, how could you express the trefoil knot in polar coordinates?

REPLY [7 votes]: When you do polar plots you are stuck with parametrizations of the limited form $$x(\varphi)=r(\varphi)\cos \varphi, \quad y(\varphi)=r(\varphi)\sin(\varphi).$$ 
The parametrization that you gave is not of that form. This is apparent already from the observation that the parametrization gives the full trefoil, when $t\in[0,2\pi]$, but the trefoil wraps around the origin twice, so $\varphi$ should range over $[0,4\pi]$. 
Let us look at the Wikipedia parametrization of the trefoil on the surface of a torus:
$$
x=(2+\cos3t)\cos2t,\quad y=(2+\cos3t)\sin2t,\quad z=\sin 3t.
$$
If we ignore that $z$-coordinate for a moment, we see $(x,y)\uparrow\uparrow(\cos 2t,\sin 2t)$, which is a tell-tale sign that here $\varphi=2t$ is the polar angle coordinate. As $t$ ranges over $[0,2\pi]$, we should, indeed, have $\varphi\in[0,4\pi]$. Thus the projection of that trefoil onto the $xy$-plane comes from the polar equation
$$
r=2+\cos\frac{3\varphi}2,
$$
as suggested by heropup (+1). The plot is not quite what you may have expected:

Here the additive constant $2$ represents the ratio of the radius of the "wire" inside the torus to that of the "tube" around the wire. IMVHO the projection looks a bit cleaner, if we use ratio $4$ and equation $r=4+\cos\frac{3\varphi}2$ instead:

For a better view here is a 3D-image of how the trefoil wraps itself around the torus.

The trefoil is the thin tube on the surface of the doughnut.