Polar to cartesian form of r=cos(2θ) This is possible for $r=\sin(2θ)$: Polar to cartesian form of $ r = \sin(2\theta)$
Surely there is some trig identity that may substitute for $cos(2θ)$ and allow for a similar coordinates transfer. What is the cartesian form of $\cos(2\theta)$?
I found something remotely similar: 
$$\cos(2θ) = \cos^2θ − \sin^2θ = 2 \cos^2θ − 1 = 1 − 2 \sin^2θ$$
(source: http://www.math.ups.edu/~martinj/courses/fall2005/m122/122factsheet.pdf)
However they all use a squared form of sin or cos, which I am not certain how to convert into Cartesian coordinates. 

All polar to Cartesian / Cartesian to polar transformations derive from these simple rules
$r^2 = x^2 + y^2\\
x = r \cos \theta\\
y = r \sin \theta$
$r  = cos 2\theta$  the 4 petaled rose.  If it had an elegant form in Cartesian we would teach it.  It will likely be a cubic or quartic equation.
$r  = \cos^2 \theta - sin^2 \theta\\
r^3 = r^2 \cos^2 \theta - r^2 \sin^2 \theta\\
(x^2+y^2)^{\frac32} = x^2 - y^2\\
(x^2 + y^2)^3 = (x^2 - y^2)^2$
I suppose we could keep multiplying that out.  But I think that looks pretty elegant. 
