# The Fourier Series of Sin^5(x)

1. Mar 10, 2008

### castusalbuscor

1. The problem statement, all variables and given/known data

So I have to find the fourier series for $$sin^{5}(x)$$.

2. Relevant equations

I know the $$a_{n}$$ in:
$$\frac{a_{0}}{2} + \sum^{\infty}_{n=1}a_{n}cos_{n}x + \sum^{\infty}_{n=1}b_{n}sin_{n}x$$
goes to zero, which leaves me with taking the $$b_{n}$$.

3. The attempt at a solution

So what I got so far is trying to integrate to find $$b_{n}$$.

$$b_{n} = \frac{1}{\pi} \int^{\pi}_{-\pi} sin^{5}(x)sin(nx)$$

But I am not sure how to proceed from here, do I make use of $$sin^{2}(x)=1/2(1-cos2x)$$ and $$cos^{2}(x)=1/2(1+sin2x)$$?

Am I even going in the right direction?

edit:

I just plugged it into Maple and got:

$$\frac{1}{\pi}\left( \frac{sin((n-5)x}{32(n-5)} - \frac{sin((n+5)x}{32(n+5)} - \frac{5sin((n-3)x}{32(n-3)} + \frac{5sin((n-1)x}{16(n-1)} - \frac{5sin((n+1)x}{16(n+1)} \right)^{\pi}_{-\pi}$$

is this the direction I need to go in?

Last edited: Mar 10, 2008
2. Mar 11, 2008

### notmuch

Euler's Formula

I did not attempt this, but have you tried using Euler's formula to help simplify before plugging in the series?

3. Mar 11, 2008

### HallsofIvy

Staff Emeritus
Have you tried using trig identies to write sin5 in terms of cos(nx) and sin(nx)?

For example, Since cos(2x)= cos2(x)- sin2[/sup= (1- sin(x))- sin2(x)= 1- 2sin2(x), sin2(x)= (1/2)(1- cos(2x)), sin5(x)= (sin2(x))(sin2(x))(sin(x))= (1/2)(1- cos(2x))(1/2)(1- cos(2x))(sin(x))= (1/4)sin(x)- (1/2)cos(2x)sin(x)+ (1/4)cos2(x).
Replacing sin2(x) in the first identity with 1- cos2(x) will give you cos2(x)= (1/2)(1+ cos(2x)) and you can use sin(a+ b)= cos(a)sin(b)+ sin(a)cos(b), sin(a-b)= -cos(a)sin(b)+ sin(a)cos(b) so that, adding the two equations. 2cos(a)sin(b)= sin(a+b)+ sin(a-b) with a= 2x, b= x, to resolve that cos(2x)sin(x). When you have reduced all products to sin(nx) and cos(nx), you have your Fourier series.

4. Mar 11, 2008

### castusalbuscor

Solved?

I ended up going with
$$\frac{1}{\pi}\left( \frac{sin((n-5)x}{32(n-5)} - \frac{sin((n+5)x}{32(n+5)} - \frac{5sin((n-3)x}{32(n-3)} + \frac{5sin((n-1)x}{16(n-1)} - \frac{5sin((n+1)x}{16(n+1)} \right)^{\pi}_{-\pi}$$

and I just plugged in $\pi$ and $-\pi$. I talked to one of the "smart" kids in class about it and I was on the right track, which was quite a relief.

Thanks for your help!

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