The Fourier Series of Sin^5(x)

[castusalbuscor]

Homework Statement

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

Homework 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}.

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?

 

[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?

 

[HallsofIvy]

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

For example, Since cos(2x)= cos²(x)- sin^2[/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.}

 

[castusalbuscor]

Solved?

I ended up going with
<br /> \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}<br />

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!