A Fourier series is an infinite series involving terms in sin(nx) and cos(nx). Since one part of the function you're working with is 1/2 cos(2x), you're halfway home to your answer, meaning that the Fourier series for 1/2 cos(2x) is just 1/2 cos(2x). All you need to do for your problem is to find the Fourier series for 1/2.I am having a difficult time finding the fourier series for (sin x)^2.
I am aware that I have to use the trig identity:
(sin x)^2=1/2-1/2(cos 2x) but I don't know where to go from here. ANy help would be greatly appreciated thanks!!
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