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

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SUMMARY

The Fourier series representation of the function f(x) = sin^2(x) is derived as (1/2) - (cos(2x)/2). The calculation for the constant term a(0) yields a value of 1/2, while the coefficients b(n) are zero due to the even nature of sin^2(x). The formula for a(n) involves integrating sin^2(x) multiplied by cos(n*x) over the interval from -π to π, leading to a complex expression that simplifies to the stated Fourier series. The user encountered issues with obtaining a zero result, indicating a potential error in the evaluation process.

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  • Knowledge of trigonometric identities, specifically sin^2(x)
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Homework Statement



f(x)=sin^2(x)

Homework Equations





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.
 
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Make sure you're not dividing by zero.
 

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