So, I was trying to find the polar equation for a sine curve and here's what I did :

Figure 1

Now, some assumptions that I've made here are : $\theta\in\left(0,\dfrac\pi2\right)$ and $C_x<\pi$, where $C_x$ is the abscissa of $C$.

So, $\sin\theta=\dfrac{BC}{r}\implies BC=r\cdot\sin\theta$ and $\cos\theta=\dfrac{AB}{r}\implies AB=r\cdot\cos\theta$. Also, if the length of $AB$ is $k$ units, then


which is the trigonometric equation that we need to solve.

My knowledge of trigonometry is only limited to 11th grade and so far, I haven't been able to think of any way to approach this equation.

Now, I don't have any formal education regarding polar coordinates and I am just trying to do this for fun, which means that my belief that solving for $r$ in terms of $\theta$ will give the equation of the sine curve when plotted in the polar plane might be wrong but I'm still fairly confident that at least the part of the sine wave lying in the interval $(0,\pi)$ will be plotted.

I would greatly appreciate any help and hints, thank you!

  • 4
    $\begingroup$ Since $r$ is both "inside" and "outside" of a trig function, the equation is transcendental, with no elementary closed-form solution. $\endgroup$
    – Blue
    Jan 10, 2021 at 11:17
  • $\begingroup$ @Blue I don't completely understand what a transcendental equation means yet but does this mean that a polar equation for the sine curve does not exist? $\endgroup$ Jan 10, 2021 at 12:16
  • 3
    $\begingroup$ "does this mean that a polar equation for the sine curve does not exist?" ... No, it means that an explicit polar equation of the form $r=f(\theta)$ does not exist; that is, there's no "formula" to generate $r$-values from $\theta$-values. Even so, the implicit form $\sin(r\cos\theta)=r\sin\theta$ counts as the "polar equation for the sine curve". It's just that if we want to get at the $r$-value(s) corresponding to a particular $\theta$-value, we have to resort to numerical methods to approximate those values. $\endgroup$
    – Blue
    Jan 10, 2021 at 19:08
  • 1
    $\begingroup$ @Blue I see, thanks! $\endgroup$ Jan 10, 2021 at 19:45


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