A circle's sine wave is an ellipse's... We all know what a sine wave is, and how it relates to a circle. 
What is the vertical and horizontal distance when I take a point and drag it along the perimeter of the ellipse?
It definitely has to be a sinusoidal function, as an ellipse is a closed conic, much like the circle... 
 A: If you drag your point so that the angle to the origin increases at a constant rate, then @wltrup's comment gives you the answer. 
If, on the other hand, you drag it at a constant speed (as if you were in a car driving along the ellipse at a steady 30 MPH), then $x$ and $y$, as functions of time, are fairly complicated. Note that if you're on a circle, "constant speed" and "constant angle change" are the same concept, so it's not clear, from your question, which one you wanted generalized. 
If, instead of driving on an ellipse, you were driving on a leminscate (sort of a figure-8 kind of curve), it'd turn out that it's pretty difficult to write down the position as a function of time. But there's a remarkable thing: your position at time $2t$, is closely related to your position at time $t$. For the circle, the corresponding question would be "are the sine and cosine of $2\theta$ related to the sine and cosine of $\theta$?", and the answer, as you may know, is this:
Yes, they are, for
$$
\sin(2\theta) = 2 \sin \theta \cos \theta\\
\cos(2 \theta) = \cos^2 \theta - \sin^2 \theta,
$$
i.e., the coordinates of the second point can be written as polynomial expressions in the coordinates of the first point. 
For the lemniscate, it turns out that the coords of the second point can be written as rational functions of the coords of the first point. Fagniano discovered this, told Euler about it, and Euler generalized out of all proportion and more or less invented Algebraic Geometry. (Or at least that's the story I got from a professor at Princeton about a million years ago when I was a ugrad there.) 
So: your very simple question, or something quite like it, was the source of a major development in mathematics a few hundred years ago. Pretty neat, eh? 
A: Since I came here in search of a steeper sine wave, I'll answer that:
y=sin((pi/2) sin(x))

That one was a bit squarish for my needs, I found you can keep a nice curve on it and still bring it back to the -1 and 1 boundaries by nesting sine waves within each other.
sin(sin(sin(x))) / sin(sin(1))

The logic I eventually figured out was this: we want the angle to change at a faster rate when it's not near pi/2 and 3pi/2, which is what a sine-wave does. So, if we did sin(sin(x)), that would cause it to move faster near the x axis. But that also means it only has values between -1 and 1, and and sin(1) doesn't get us to pi/2, so the height of our sine wave is diminished. It will never be higher than sin(1)~0.84, so we divide by sin(1) to stretch it vertically so that it's previous highest and lowest values get stretched to 1.
