How from Fourier Transform with imaginary numbers we get the one with real numbers?

Question

There is Fourier Transform formula with e pow i, there is also another Fourier Transform formula with real numbers only (with sin and cos).

I thought that imaginary numbers are imaginary only except for e pow 2pi which is 0 and e pow pi which is 1.

How come that imaginary numbers become all of a sudden real numbers?

Answer 1

Because we have that $$\sin{(x)}=\frac{e^{ix}-e^{-ix}}{2i}$$ $$\cos{(x)}=\frac{e^{ix}+e^{-ix}}{2}$$ So the complex exponentials can be rewritten as sine and cosine.

Answer 2

After some research I've edited the Hebrew wiki site page, but since I'm not native English speaker, here's an explanation in-here:

As Peter, who has answered the question said, there is a formula that gives and provides for each Fourier in ${C}$, it's result in ${R}$. But in order to understand who's he do that, we'll need first to understand how's the fourier function being built.

Recall that -

${e^i}$ is a circle with radius ${1}$ .

Taking periodic ${f(x)}$ and having:

${e^{i{\omega}f(x)}}$

Would lead us to a function drawn 'over' circle with radius 1 whereas each point on the circle would give us the point on the function, while the distance of ${f(t)}$ on the circle from the middle of the circle, is the height of ${f(t)}$ from ${Xt}$ axis.

So to answer my own question, getting real numbers from $C$ dimension functions, is done by calculating distance (which is $R$), from the middle of the circle (which is simple expressed using - not in - using different dimensions).