Fourier Series for Even Functions

For anevenfunctionf(t), defined over the range -LtoL(i.e. period = 2L), we have the following handy short cut.

Since

and

it means the integral will have value 0. (See Properties of Sine and Cosine Graphs.)f(t) is even,

So for the Fourier Series for an even function, the coefficientbhas zero value:n

So we only need to calculatebn= 0a0 andanwhen finding the Fourier Series expansion for an even functionf(t):

Anevenfunction has onlycosineterms in its Fourier expansion:

Specally how he has written ao/2 in last equation. Thanks