Why is there a Half Cos(x) Term in the Fourier Series for Cos(x)?

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SUMMARY

The discussion focuses on the Fourier series representation of the function f(x) = Cos(x) defined on the interval [-π/2, π/2] and zero elsewhere, with a period of 2π. The key point is the presence of the half Cos(x) term in the Fourier series, which arises from the specific integration limits affecting the coefficients. The user confirms that the Bn coefficients equal zero due to the orthogonality of sine and cosine functions, which holds true when integrating over the specified interval.

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Hello, I'm having some trouble completing a Fourier series question.

We have f(x)=Cos(x), for abs(x)≤π/2
f(x)=0, for π/2≤abs(x)≤π
period=2π

and then show that the Fourier series f(x) reads
http://img71.imageshack.us/img71/3424/84335314oi1.jpg
now I can get the 1/π and the sum, as an=
http://img71.imageshack.us/img71/1703/eq2hn6.jpg
and working through from there, the only part I'm having trouble with is the origin of the half cos(x) term. As far as I'm able to apply the Fourier series I don't see where it comes from. (Bn must equal 0 right?)
 
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How did you deduce, that the Bn are zero? The orthogonality relation between sin and cos only holds if you integrate from 0 to 2pi, but in this case you integrate only from 0 to pi/2.
 

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