Fourier series for $\sin^2(x)$ I was asked to compute the Fourier series for $\sin^2(x)$ on $[0,\pi]$. Now this is what I did and I'd like to know if I'm right.  $\sin^2(x)=\frac12-\frac12\cos(2x)$ . I got the right hand side using trig identities. I'm wondering If I can do this without using the formulas. Thanks.
One small point...

The way the question is stated, there may be a slight ambiguity.  One way (and almost certainly the intended way) to read the question is:  given the (periodic) function $\sin^2(x)$, find its Fourier series on the interval $[0, \pi]$.  In this case, $(1 - \cos(2x))/2$ is correct.

However, we could also read it as follows:  given the function $\sin^2(x)$ defined on the interval $[0, \pi]$, find its Fourier series.  In this case, we must first decide how to extend the function to be periodic.  Of course the natural choice is to extend it to equal $\sin^2(x)$ for all $x$.  Again, the cosine series is correct.

But... we do have the freedom to extend $\sin^2(x)$ to an odd function on $[-\pi, \pi]$ instead, in which case the Fourier series will contain only sine functions (with the coefficients computed in the usual way). The point being that there is in fact another series, featuring only sines, that converges to $\sin^2(x)$ on the interval $[0, \pi]$.  Of course, the convergence isn't as fast ;-).
honestly, i don't think that $b_n = 0$ people. unless somebody can convince because the integral of $sin(x)^2 * sin(nx) dx$ is  not zero on $(0, \pi ) $. it actually has a value and only zero when n = 2. just saying. the post mentioned this interval, i just wondered why 
most of you assumed $(-\pi, \pi)$  if this is still nonsense to you, you might as well delete this post too, mr.modulator.