Why must the coordinates of a point on a circle be sinusoidal as a function of the angle?

Question

A defining feature of radians as a unit of measurement is that if an angle $\theta$ is expressed in radians, the height of a point on the unit circle at this angle is $\sin(\theta)$.

Is it possible to choose a different unit of measurement for the angle so that we get something other than a sinusoid?

Why doesn't the height of a point on the unit circle as a function of the angle give us a well-shaped half-circle on each half-period rather than a sinusoid? Wouldn't such a circle shape be more fundamental than a sinusoid?

If not, why must the coordinates of a point on the unit circle be sinusoidal as a function of the angle?

I may have a very basic error in my thought, maybe sines were not taught properly to me, but when I learned them they were something that suddenly came out of nowhere. I know it's like the movement of a wriggling snake or the time diagram of a spring that is bouncing, but I still have this question in my head. I would be thankful if I hear your input on it.

Answer 1

Suppose $\theta$ is an angle in radians. Let's create a new unit for angles called the $\mathrm{Moytaba}$, or $\mathrm{Moy}$ for short. Let us define it by $1 \,\,\mathrm{Moy} = c\,\,\mathrm{rad}$ where $c$ is some constant scaling factor (of course, any transformation between units of the same dimension has to be a constant scaling factor, for obvious reasons).

Then we know that $\theta \,\,\mathrm{rad}$ is just $\frac{\theta}{c} \,\,\mathrm{Moy}$. Since the height of a point on a circle wrt the angle in radians gives us a height of $\sin(\theta)$, then the height of a point on the circle in $\mathrm{Moy}$s must be $\sin(\frac{\theta}{c})$. Clearly this is still sinusoidal! So this is your proof that any other units are still sinusoidal (and just represent scaling the period of the sine wave).

But for the real intuitions, as I noted in my comments above:

I think this animated GIF (which handles the angle in a purely unit-less fashion) will give you the intuition you're after: http://en.wikipedia.org/wiki/File:Circle_cos_sin.gif

and

In fact, the reason why sine describes the time diagram of a bouncing spring is because the Kinetic E + Potential E = Total Energy (constant) equation actually parametrises a circle, where we end up getting momentum/velocity along one axis and position along the other. As time passes, we just move around this circle, with energy flowing back and forth between kinetic and potential. When you graph the position with respect to time, you're just getting the height of the point on the circle, just like we are here! http://en.wikipedia.org/wiki/File:Simple_Harmonic_Motion_Orbit.gif

Edit: After talking a little more with the OP, I couldn’t resist citing one last, brilliant animation: https://www.deviantart.com/woodmath/art/Euler-s-formula-3d-visualization-268936785

Answer 2

NOTICE: This answer is NOT CORRECT. It's only not deleted because it might be a useful mistake to inform unaware people that sinusoids and cycloids are not the same things. what the answer below describes is indeed a cycloid and NOT a sinusoid.

We can rotate the circle itself too! Say it's a wheel and there's a marker on some point of its circumference and this wheel is rotating near a wall(so that the marker will draw the path taken by the point for us), now the change in the angle between the horizontal radius of the circle and that point is what we simulate by rotation of the wheel!.

I've drawn different states of the wheel in the picture below, we can see that even if the circle moves an epsilon(assuming epsilon is an infinitesimally small number) forward, the new location of the point can not be on the same previous circle obviously, and that means it's impossible to produce a movement path that consists of half circles for a fixed point on a rotating circle. Instead, they will be in a shape which we call "sinusoidal"(they are indeed cycloids).

Moving Circle With Fixed Point On It