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?
This question is of interest to me, too. I do not have an answer to the question above, but can provide some additional information. First, if one enters sqrt((sint+2sin2t)2+(cost+2cos2t)2) into Wolfram Alpha, it returns a different derivation, which when plotted does not look anything like sqrt(5−4cos3t). Second, I think the problem with the logic above is that some information must be lost in the process of squaring the statements. By playing around in gnuplot, the closest I've been able to get to a knot is 1+0.05cos(3t+0.5t) over -2pi..2pi, 1+0.05cos(3t+0.3t) over -3pi..3pi. One can play with the scaling of t as desired (primes just as in knot theory?) but the range has to be adjusted accordingly. I hope someone else can put this in the context of the original question. Regards, David