# Closed form of a summation

1. Jan 14, 2008

### Frillth

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

I am looking for a closed form of the summation:
sin(x) + sin(3x) + sin(5x) + ... + sin((2n-1*)x)

2. Relevant equations

None.

3. The attempt at a solution

Through a complete stroke of luck, I believe I have arrived at the correct solution: sin^2(nx)/sin(x)
I have tested this for many different cases, and I believe it is correct. However, I am having a hard time proving that it is. Can anybody point me in the right direction?

2. Jan 14, 2008

### rock.freak667

How did you prove that?

[EDIT: The last term should be sin(2n+1)x not 2n-1

3. Jan 14, 2008

### Hurkyl

Staff Emeritus
Seems to me that mathematical induction would be an obvious thing to try...

4. Jan 14, 2008

### Frillth

I'm a little rusty on my induction skills. Is this what I need to do?

1. Show that:
sin^2(nx)/sin(x) + sin((2(n+1)-1)x) = sin^2((n+1)x)/sin(x)

2. Show that my formula works for any specific case.

5. Jan 14, 2008

### rock.freak667

You need to show that

$$\sum_{n=0} ^N sin(2n-1)x= \frac{sin^2(Nx)}{sinx}$$

then add the (N+1)th term to each side and show that is can be written as

$$\frac{sin^2((N+1)x)}{sinx}$$

Last edited: Jan 14, 2008
6. Jan 14, 2008

### Frillth

I've been trying to get these two sides equal, but I'm not coming up with anything. Which identities should I be using to tackle this problem?