Converting Polar Equation to Cartesian Equation: general form solution

Question

I'm trying to find the Cartesian equivalent of the general equation $$r=a\cos(q\theta) + c; q\in\mathbb Q, a\gt c \in\mathbb R$$ if it exists. My memory of calc is a bit hazy, and I haven't been able to break this up using any of the trig identities that I remember, or could find online. I've also tried using Laplace transforms for fixed $a, c, q$. However, this resulted in an expression containing complex exponentials. For example $$r(\theta)=\cos\left(\frac 57\theta\right) + 2=\frac 1 {98}\left[e^{-5i\theta}+e^{5i\theta}\right] + 2$$ This strategy wasn't very fruitful as I have almost no background in complex math, and have no idea how to handle those exponential terms.

I'm very curious to know weather there's a general cartesian solution corresponding to this type of polar equation. I'd also be exceedingly grateful for a worked example using set constants, or recommendations for reading material that might help me to figure this out.

Answer 1

You can write this as $$\sqrt{x^2+y^2}=a\cos\left(q\arctan(\frac yx)\right)+c$$

Answer 2

If you restricted the solution set to $q$ being an integer, then you could make use of this identity: $$ \cos(q\theta)=\sum_{k=0}^q\cos^k(\theta)\sin^{q-k}(\theta)\cos\left(\frac{(q-k)\pi}{2}\right) $$ Then using: $$\begin{align} r&=\sqrt{x^2+y^2}\\ \cos(\theta)&=\frac{x}{\sqrt{x^2+y^2}}\\ \sin(\theta)&=\frac{y}{\sqrt{x^2+y^2}} \end{align}$$ You end up with: $$ \sqrt{x^2+y^2}=a\left(\sum_{k=0}^q\frac{x^ky^{q-k}}{(x^2+y^2)^\frac{q}{2}}\cos\left(\frac{(q-k)\pi}{2}\right)\right)+c $$ (assuming I haven't made any errors in my calculations)