Usually, applying the conversion formulae
$r^2=x^2+y^2$
$\cos\;\theta=\frac{x}{r}$
$\sin\;\theta=\frac{y}{r}$
to transform an equation in polar coordinates to an implicit Cartesian equation is quite straightforward for an equation of the form
$r=f(\cos(n\theta),\sin(n\theta))$
with $n$ an integer, thanks to multiple angle formulae. Polar equations of the form
$r=f\left(\cos\left(\frac{p}{q}\theta\right),\sin\left(\frac{r}{s}\theta\right)\right)$
where $p$, $q$, $r$ and $s$ are integers, and $\frac{p}{q}$ and $\frac{r}{s}$ are in lowest terms, are more difficult to handle, but the case where $q$, $s$ are powers of 2 is slightly eased by the existence of identities like $\tan\left(\frac{\theta}{2}\right)=\frac{\sin\;\theta}{1+\cos\;\theta}$.
The really difficult ones to handle are the cases like Cayley's sextic:
$r=4a\cos^3\left(\frac{\theta}{3}\right)$
and in general, the cases with fractions whose denominators are not powers of 2. In particular for Cayley's sextic, I can't seem to find a quick way to exploit the appropriate multiple angle identities.
One "cheat" I have seen is to instead represent the polar equation as a pair of parametric equations, and then make the substitution $\theta=3\arctan\;t$ so that everything is represented algebraically. The problem is that apparently it takes some insight to recognize how to remove the new parameter $t$ easily. For even tougher cases like
$r=\cos\left(\frac{2\theta}{3}\right)-\sin\left(\frac{3\theta}{5}\right)$
the appropriate substitution after transforming to parametric equations is $\theta=\mathrm{LCM}(3,5)\arctan\;t$, but the expressions, though algebraic, look even more nightmare-ish to manipulate.
Mathematica has no trouble figuring out the implicit Cartesian equation, through a judicious use of Gröbner bases:
Factor /@ GroebnerBasis[Append[Thread[{x, y} == TrigExpand[{4a Cos[t/3]^3 Cos[t], 4a Cos[t/3]^3 Sin[t]}]],
Cos[t/3]^2 + Sin[t/3]^2 == 1], {x, y}, {Cos[t/3], Sin[t/3]}, MonomialOrder -> EliminationOrder]
but I'm pretty sure the Cartesian equations have already been determined way before Buchberger was born.
How would one determine the Cartesian equation of such a curve using only classical techniques?
Apropos to this question, is there a quick way to determine the degree of an algebraic curve represented in polar coordinates without having to do a conversion?
Best Answer
We could do something simple, since we have: \begin{equation} \cos\left(\frac{p}{q}\theta\right)=\frac{1}{2}\left(e^{i\frac{p}{q}\theta}+e^{-i\frac{p}{q}\theta}\right)=\frac{1}{2}\left(\left(\frac{x}{r}+i\frac{y}{r}\right)^{\frac{p}{q}}+\left(\frac{x}{r}-i\frac{y}{r}\right)^{\frac{p}{q}}\right) \end{equation} That is \begin{equation} \cos\left(\frac{p}{q}\theta\right)=\frac{1}{2}\left(\frac{1}{x^2+y^2}\right)^{\frac{p}{2q}}\left(\left(x+iy\right)^{\frac{p}{q}}+\left(x-iy\right)^{\frac{p}{q}}\right) \end{equation} and similarly we have \begin{equation} \sin\left(\frac{p}{q}\theta\right)=\frac{1}{2i}\left(\frac{1}{x^2+y^2}\right)^{\frac{p}{2q}}\left(\left(x+iy\right)^{\frac{p}{q}}-\left(x-iy\right)^{\frac{p}{q}}\right) \end{equation} Now we can use these to express your equations in terms of $x$ and $y$. Is this any use?