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

Hey there!
I'm trying to calculate the Fourier Series for sin2x on [-π, π]

For a0 I found 1/2. (By determining the average value of the function on the interval)

Since sin2x is even, I know that bn = 0.
Now, for an.. The following link shows the integral I used to try to evaluate an.
http://www.wolframalpha.com/...

Since an and bn are both coming out to be zero, doesn't this imply that sin2x = a0 = 1/2? Will someone please show me where I am going wrong? Can't find anything online about this. Everyone just always uses the sin2 identity and calls it good.

Last edited:

lurflurf
Homework Helper
hint: Recall that
(sin(x))^2=(1/2) (1-cos(2 x))

as for the integral we have

$$a_2=\frac{1}{\pi} \int _{- \pi}^\pi \! \sin^2(x) \cos(2x) \, \mathrm{dx}=\frac{1}{2\pi} \int_{-\pi}^\pi \! (1-\cos(2 x)) \cos(2x) \, \mathrm{dx}$$

What can I say.... My mind is blown...

When you take the integral of sin2(x)cos(nx) you get -$\frac{4sin(πn)}{n^3-4n}$

It is easy to assume that this is equal to zero for all values of n, but in hindsight it is clear that this does not apply for n = 2. :S!

I'm utterly confused now. Just when I thought I was beginning to understand how to compute Fourier Series......

mathman
What can I say.... My mind is blown...

When you take the integral of sin2(x)cos(nx) you get -$\frac{4sin(πn)}{n^3-4n}$

It is easy to assume that this is equal to zero for all values of n, but in hindsight it is clear that this does not apply for n = 2. :S!

I'm utterly confused now. Just when I thought I was beginning to understand how to compute Fourier Series......

For n = 2, use L'Hopital's rule. I got -2π.

One could also go to basics, using cos2x = cos2x-sin2x
Therefore sin2x = 1/2(1 - cos2x)

Last edited:
vanhees71
Gold Member
It's clear that $a_0=1$ and $a_1=-1/2$, because as said in the previous posting from the double-angle theorem you get
$$\sin^2 x=\frac{1}{2}[1-\cos(2 x)].$$
For the Fourier coefficients you indeed have
$$a_n=\frac{1}{\pi} \int_{-\pi}^{\pi} \mathrm{d} x \sin^2 x \cos(n x)=\frac{4}{\pi(4n-n^3)} \sin (n \pi).$$
This is indeed $0$ for all $n \in \mathbb{N} \setminus \{2 \}$, and for $n=2$ you get by direct evaluation of the integral $a_2=-1/2$.

Of course you can also take the limit $n \rightarrow 2$ by using de L'Hospital's rule, because it's a limit of the indefinite form 0/0:
$$\lim_{n \rightarrow 2} a_n=\lim_{n \rightarrow 2} \frac{4}{\pi} \frac{n \pi \cos(n \pi)}{4-3n^2}=-\frac{1}{2}.$$