Fourier series expansion. Find value of a term in expansion

[jaus tail]

Homework Statement

Fourier series expansion of a signal f(t) is given as
f(t) = summation (n = -inf to n = +inf) [3/(4+(3n pi)²) ) * e ^{j pi n t}

A term in expansion is A₀cos(6 pi )
find the value of A₀

Repeat above question for A₀ sin (6 pi t)

Homework Equations

Fourier expansion is summation n = -inf to +inf C_n e^jwnt
where C_n is Integration over T₀ x(t) e^-jwnt dt

The Attempt at a Solution

In book they've said C_n is right.
But then they say a_n is 2 times real part of C_n
and have done a_n = 6/[4 + (18 pi)²

For third part they've put A₀sin (6 pi t) = bn sin (n w t)
and for n = 6, this is zero.

I didn't understand this part. Why did they multiply it by 2 first and then how did it become zero in the second.

I understand that sin ( 6 pi) is zero but how can sin ( 6 pi t) be zero? Wouldn't this vary as 't' varies?[/B]

 

[pasmith]

The coefficients of \cos n\pi t and \sin n\pi t come not just from c_n, but also c_{-n}: <br /> c_ne^{i n\pi t} + c_{-n}e^{-in\pi t} = (c_n + c_{-n})\cos n\pi t + i(c_n - c_{-n})\sin n\pi t

 

[jaus tail]

Oh yeah. Thanks. I didn't think about C_-n.