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Showing fourier series of sin^2(x)=(1/2)-(cos(2x)/2)

1. The problem statement, all variables and given/known data

f(x)=sin^2(x)

2. Relevant equations



3. The attempt at a solution

solving for a(0)= i did (1/2Pi)*int(sin^2(x),x,-Pi..Pi)=1/2
b(n)=0 because sin^2(x) is an even function
a(n)=(1/Pi)*int(sin^2(x)cos(n*x),x,-Pi..Pi)=1/2Pi((sin(xn)/n)-(.5sin((2-n)x)/(2-n))-(.5sin((2+n)x)/(2+n))) this whole thing evaluated from -Pi..Pi

I keep getting zero although I know this is not the answer, so maybe I am messing up somewhere. Help is greatly appreciated.
 

vela

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Make sure you're not dividing by zero.
 

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