Cardioid: converting parametric form into polar coordinates I am interested in converting parametric equations:
$$\varphi=\left(\varphi_{1},\varphi_{2}\right)=\left(2\cos{t}-\cos{2t},2\sin{t}-\sin{2t}\right)$$
which describe a cardioid, into polar coordinates. What would be a good beginning for this type of an operation? I would be thankful for some hints. 
 A: The parametric equations describe $(x,y)(t) = (2 \cos(t) - \cos(2t), 2 \sin(t) - \sin(2t))$:

In order to convert this into polar coordinates, express the radius, and the angle in terms of $x$ and $y$ first:
$$
    r(t)^2 = x(t)^2 + y(t)^2
$$
this would be a simple expression in terms of $\cos(t)$. You could them express the polar angle using atan2 as follows $\theta(t) = \arctan(x(t), y(t))$. The radial representation would be obtained if it were possible to solve for $t = t(\theta)$, and substituted into $r(t)$ to obtain $r(\theta)$. But the equation is not linear and does not admit a simple closed-form.
However, seeing that $r(t)$ is much simpler, we can invert it to find $t=t(r)$ and then $\theta(r) = \theta(t(r))$. Doing so will only allow us to parametrize half of the cardoid, due explicit symmetry $r(t) = r(2\pi - t)$.
If you permit the use of software, here is a way to get the radial representation:

A: I was working to this one by very tedious manipulating relations and some substitutions in Maple that I saw this old question. I can see that we get $$(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$$ The final step indeed is to use $x=r\cos\theta,~~y=r\sin\theta$.
A: Hint: compute the radius and the angle as a function of $t$! ;-)
The radius is the square root of the sum of the squares of the two components; the angle can be determined using the inverse of the tangent function. 
A: In GeoGebra you could try something like this:
https://i.imgur.com/6GQr31V.png
There is still a minor issue with that attempt (near the points where the graph crosses the y-axis), but with a little tweaking it should work out ok.
To elaborate a little: The rectangular form of the curve is given in terms of horizontal and vertical displacement in a parameter. If you use the Pythagorean theorem, you can obtain the displacement along a ray starting at the origin. Then you use the arctan of the ratio of vertical displacement over horizontal displacement to transform the parameter in a suitable fashion. This can sometimes result in a graph that is a mirror image of the curve that you're after, hence there is an if command that applies a transformation when the parameter is in a certain range and the chosen range of the parameter can also make a difference. By tweaking those aspects, you can obtain the desired parametric curve in polar form. With Wolfram Alpha you can find the root of 2 cos(x)-cos(2x) (the point where the graph intersects the x-axis) and this can be used to set the appropriate range.
https://i.imgur.com/j5onZrr.png
https://i.imgur.com/SAxavZl.png
