Is it possible to express a cycloid in polar coordinates

Question

The parametric equations for a cycloid of radius 1 centered at the pole are $$ x(t) = t - \pi - \sin t \\ y(t) = \pm (1- \cos t) $$ where the plus sign is a cycloid above the x-axis and the minus sign a cycloid below the x-axis. Is it possible to convert these equations into polar coordinates so that the cycloid is expressed as a angle dependent radius $r(\theta)$ ? I tried to convert them to polar coordinates by using $$ r = \sqrt{x(t)^2 + y(t)^2} \\ \tan \theta = \frac{y(t)}{x(t)} $$ But I don't understand how I can eliminate the parameter $t$ so that I get a function $r(\theta)$ in the end.

Answer 1

To express $\rho$ as a function of $\theta$, you could express $t$ as a function of $\theta$ by inverting

$$\tan\theta=\pm\frac{1-\cos t}{t-\pi-\sin t}$$and plug the result in the expression of $\rho=\sqrt{x^2+y^2}$.

But there is no closed-form expression for $t(\theta)$ because $t$ appears inside and outside of a trigonometric function, making a transcendental equation.

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Following @N74's approach, you can write

$$\rho\cos\theta=\arccos(1\pm\rho\sin\theta)-\pi-\sin(\arccos(1\pm\rho\sin\theta)),$$ which is a nasty implicit polar equation.