# Fourier series for (sin(x))^2

by Gza
Tags: fourier, series, sinx2
 P: 525 Hey guys i was working on an algorithm for one of my CS classes that included working out the fourier series for the function f(x) = (sin(x))^2. it's been a few years since i've done anything like this, so I did some googling to refresh my memory of how to determine the fourier coefficients, and after some integrals and calculations, came up with: f(x) = (1/2) + (1/2)cos(2x) + (1/2)sin(2x) I graphed this along with f(x) = (sin(x))^2 and it looked different from it. I was wondering if someone can give me a correct answer to compare with what i came up with, or help with giving me a rough walkthru of the process. Thanks again.
 HW Helper P: 4,124 The answer you got is incorrect. The two identities you need are: (sin(x))^2 + (cos(x))^2 =1 and cos(2x) = 2(cos(x))^2 - 1 Use both of these to solve for (sin(x))^2 in terms of cos(2x).
HW Helper
P: 3,031
 Quote by Gza Hey guys i was working on an algorithm for one of my CS classes that included working out the fourier series for the function f(x) = (sin(x))^2. it's been a few years since i've done anything like this, so I did some googling to refresh my memory of how to determine the fourier coefficients, and after some integrals and calculations, came up with: f(x) = (1/2) + (1/2)cos(2x) + (1/2)sin(2x) I graphed this along with f(x) = (sin(x))^2 and it looked different from it. I was wondering if someone can give me a correct answer to compare with what i came up with, or help with giving me a rough walkthru of the process. Thanks again.
One thing you should always check is to see if the symmetry of the component functions matches the symmetry of the function you are decomposing. The function you started with is an even function

f(-x) = f(x)

All the component functions must also be even functions. One of yours is not. You can often take advantage of symmetry to eliminate performing many of the integrals that have to be calculated to decompose a function.

 P: 998 Fourier series for (sin(x))^2 No integrals at all required here. You can either use the identities learningphysics gave, or use the exponential form of sine: $$\sin{x} = \frac{e^{ix}-e^{-ix}}{2i}$$ and just square it (and then convert back to sines and cosines).
 Quote by Data No integrals at all required here. You can either use the identities learningphysics gave, or use the exponential form of sine: $$\sin{x} = \frac{e^{ix}-e^{-ix}}{2i}$$ and just square it (and then convert back to sines and cosines).