Derivation of the equation for A_{n}. Multiply both sides of the Fourier series equation by cos(2πmf^{ 0}t) and integrate over one repetition period, t^{0}.

When n ≠ m, all terms on the right side integrate to zero. The proof of this assertion involves the trig identities 2cosαcosβ = cos(α+β) + cos(α-β), etc. When n = m, only the cosine product has a non-zero integral.

The zero frequency term. When n = m = 0, the first term on the right of the top equation integrates to A_{0}t/2. When added to the integral of the cosine product, the result is A_{0}t. This result can be used to compute A_{0}.

A0=1t0∫0t0F(t)cos(0)dt

This result (the correct answer) is half that obtained by the general formula for A_{m}. This explains why, in the Fourier series expansion, the value of A_{0} is divided by 2.

Derivation of the equation for B_{n}. Multiply both sides of the Fourier series equation by sin(2πmf^{ 0}t) and integrate over one repetition period, t^{0}. The derivation follows that show above for A_{n}. That is, all terms will integrate to zero except that with the product of two sines.