Prove Orthogonality Condition For Sines (Integral)

[maherelharake]

Homework Statement

I need to prove the equation attached. I also have to describe why the integrals vanish.

Homework Equations

The Attempt at a Solution

I am not sure how to begin. Our teacher told us this equation is known as the orthogonality condition for sines. I also know that n' is a positive integer. I think if I get help starting this I will be able to proceed.

 

[╔(σ_σ)╝]

Use a trig identity to convert the product to sums.

 

[maherelharake]

I worked through the integral, but I kept getting zero for all cases. I have attached my work. Any help is appreciated.

 

[╔(σ_σ)╝]

If n=n' you simply have sine squared blah blah... If you integrate it, you will not get zero. This can be intuitively seen from the fact that the graph of sine squared is strictly greater than or equal zero.

If you keep getting zero then something is wrong.

 

[maherelharake]

Right I understand that completely. I just can't seem to find my mistake. I went over it several times

 

[╔(σ_σ)╝]

For the case n=n' don't convert the product to a sum. Just integrate sine square whatever by using the half angle identity for cosine.

EDIT
Double angle not half angle.

 

[maherelharake]

But in order to prove that, don't I have to integrate regardless? I feel like it's not the right way to do it if I use the idea of n=n' beforehand.

 

[╔(σ_σ)╝]

But in order to prove that, don't I have to integrate regardless? I feel like it's not the right way to do it if I use the idea of n=n' beforehand.

What do you mean ? The question asked you to verify the formula for the two cases. So I see nothing wrong with considering the cases separatly.

What you have done ,thus far, is valid when n'=/=n.

If you look at your solution, when you converted the product to a sum if you let n=n' one of the cosine terms vanish before you integrate. You cannot just plug in n=n' into your final solution because of division by zero :-). Those constants you have in the final step or your solution become infinity when n=n'.

So you have to consider the cases separatly.

 

[maherelharake]

Hmm I see what you are saying. Well I suppose I will ask the professor if this method is acceptable, and let you know tomorrow. Thanks.

 

[╔(σ_σ)╝]

What exactly makes it seem invalid to you?

 

[maherelharake]

It just kind of seems to me that the teacher phrased the question in a way such that he wants us to completely simplify the left hand side of the equation before we use any of the knowledge on the left. I hope that makes sense in the way I explained it.

 

[╔(σ_σ)╝]

That is what we are doing.
The only thing is that if n=n' the method you are using won't work directly.

The reason is what I already pointed out. In your result of your integration you have a 1/(blah(n-n')). If you make n=n' you have a problem with dividing by zero.

This is why we have to consider different cases.

 

[maherelharake]

Ohhhh I think I see what you are saying now. Ok I will work on it and post what I have done either tonight or tomorrow. Thanks.

 

[╔(σ_σ)╝]

No problem :-). Take your time.

 

[maherelharake]

Got it now. Thanks.

 

[nareto]

aargghh why haven't you posted the solution

EDIT: nevermind, I found it here: http://www.mathreference.com/la-xf-four,orth.html and I'm copying it below for reference:

Consider two functions f = sin(mx) and g = sin(nx). Here m and n are distinct positive integers. What is the integral of f×g, as x runs from 0 to 2π?

Use the angle addition formula to write the following trig identity.

sin(mx)×sin(nx) = ½ ( cos(mx-nx) - cos(mx+nx) )

Integrate the right side from 0 to 2π and get 0. Here are two more identities to demonstrate the orthogonality of cos(mx)×cos(nx) and sin(mx)×cos(nx).

cos(mx)×cos(nx) = ½ ( cos(mx-nx) + cos(mx+nx) )

sin(mx)×cos(nx) = ½ ( sin(mx+nx) + sin(mx-nx) )