# Varification needed for small trigonometrical Fourier series, PRESSING

1. Nov 24, 2012

### toneboy1

Hi,
I have x(t) = 1/2 + cos(t) + cos(2t)

so I can see that a0 = 1/2
and that it is an even function so there is no bn
Also that T = 2pi so
an = 2/2pi ∫02pi x(t).cos(nω0t) dt

but when I integrate this I get an = 0 yet I've been told that the answer is

x(t) = 1/2 + Ʃn = 12 cos(nω0t)

which would mean that an = 1, but I'm not sure how.

Anyone?

Thanks heaps!

EDIT: I just found another related problem I'm struggling with if anyone's interested:

Last edited: Nov 25, 2012
2. Nov 26, 2012