How do all points $(x; \sin x)$ look in polar coordinate system

Question

Recently, I wondered how $(x; \sin x)$ would look in polar coordinate system (where the first coordinate is the distance from the origin, and the second one is the angle from the horizontal axis). My imagination tells me, that it should still look, like some kind of a wave, but I'm not sure. So how does it look? And, btw, is there some kind of an app to build graphs in polar? I tried in GeoGebra, but it didn't work.

Answer 1

GeoGebra works, but you have to actually use the Greek letter $\theta$ as your free variable. That's the easiest way to make GeoGebra understand that you want it to think polar.

Usually, you would let the angle be the free variable, take the sine of the angle, and let that be the distance from the origin. It turns out that the resulting graph of the sine function, in polar coordinates, is a circle with radius $\frac 12$ centered at $(0, \frac12)$. This is what appeared when I entered sin(θ):

You seem to want the opposite, though. To do that, I looked at the definition of the a object, and saw that it was defined as Curve((sin(θ);θ), θ, 0, 2π) We want to flip the two coordinates, so that's what we do: Curve((θ;sin(θ)), θ, 0, 2π) gives the curve

Of course, at this point, $\theta$ is not the correct variable name any more. It would be more correct to use Curve((r;sin(r)), r, 0, 2π). Now that we have figured out the actual syntax for making GeoGebra plot in polar coordinates, we don't need to use $\theta$ any more to force it.

Answer 2

Here is the plotting in MATLAB