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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 b_{n}

Also that T = 2pi so

a_{n}= 2/2pi ∫_{0}^{2pi}x(t).cos(nω_{0}t) dt

but when I integrate this I get a_{n}= 0 yet I've been told that the answer is

x(t) = 1/2 + Ʃ_{n = 1}^{2}cos(nω_{0}t)

which would mean that a_{n}= 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:

https://www.physicsforums.com/showthread.php?t=654607

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# Varification needed for small trigonometrical Fourier series, PRESSING

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