Fourier series (very basic question)

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I have to use some basic knowledge about Fourier series in some physics exercise, so Im trying to learn the basics, but I cant understand something that I guess is pretty basic:

given a functio $ f $ that can be represented as a Fourier series, we can write:

$ f\left(x\right)=\sum_{n=-\infty}^{\infty}C_{n}e^{inx} $

where $ C_{n}=\frac{1}{T}\int_{0}^{T}f\left(t\right)e^{-i\omega_{n}t}dt $.

and $ \omega_{n}=\frac{2\pi n}{T} $

But we could also represent $ f\left(t\right)=\sum_{n=0}^{\infty}\left(A_{n}\cos\left(\omega_{n}t\right)+B_{n}\sin\left(\omega_{n}t\right)\right) $

Where $$ A_{n}=\frac{1}{T}\int_{0}^{T}f\left(t\right)\cos\left(\omega_{n}t\right)dt\thinspace\thinspace\thinspace\thinspace,\thinspace\thinspace\thinspace\thinspace B_{n}=\frac{1}{T}\int_{0}^{T}f\left(t\right)\sin\left(\omega_{n}t\right)dt $$

But I cannot see how can it be that those forms equivalent. I'll be glad for a clarification (please consider the mose simple explanation you can think of, because I have really learned this level of calculus, Im just suppose to learn some basic formula's in order to use them in physics exercises).

Thanks in advance

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Using $e^{iz}=\cos{z}+i\sin{z}$ and the fact that $\cos$ and $\sin$ are respectively an even and an odd function, you can show that $C_n=A_n-iB_n$ and (I suppose you meant $\omega_n$ in the first formula too) $C_ne^{i\omega_nx}=(A_n-iB_n)(\cos{\omega_nx}+i\sin{\omega_nx})=(A_n\cos{\omega_nx}+B_n\sin{\omega_nx})+i(\dots)$ and the imaginary part must vanish (assuming $f$ is a real valued function).