I have $f(x) = \sin(x)$. I thought that when I do Fourier Transform and construct epicycles, than those epicycles will draw that $\sin(x)$ function (but this is probably not case with $\sin(x)$, cause there will be just one circle? So what those epicycles actually draw? I read many times that they can draw whatever curve, so I thought it draws that function from which I did Fourier Transform.
What i think is:
$f(t) = \sin(t)$ fourier transform will be..
just $frequency = 1$ with $amplitude = 1$,
so $F(1) = 1$ and $F(x) = 0$ otherwise
$F(1) = 1$ means one unit circle when constructing epicycles, but this epicycle will not 100% draw $\sin(x)$ function, cause it is just one circle.
Just $\sin t$ can't give you a circle, you need two coordinates for a point in the plane. For example, $(\cos t,\sin t)$, or in complex notation $e^{it}=\cos t+i\sin t$, draws a circle as $t$ runs over $[0,2\pi]$. If you want to approximate the graph of $y=\sin x$ with "epicycles" you need to approximate the curve $(t,\sin t)$, or $t+i\sin t$ in complex notation, with linear combinations of $\dots,e^{-it},1,e^{it},e^{2it},\dots$. Such combinations correspond to tracing rotating circles, whose centers trace other rotating circles, and their centers trace other, and so on.
Basically, you need the Fourier series of the complex valued function $t+i\sin t$. Since Fourier series converges to the function that generated it (say if the function is continuous) those epicycles will draw a piece of "whatever curve" with prescribed accuracy by cutting off the series to a finite sum. They can not approximate the whole graph at once though, because circles close and the graph of $\sin$ does not.