# Integral Product of Cosines to Show Orthogonality

1. Oct 7, 2012

### royblaze

1. The problem statement, all variables and given/known data
Show that (forgive me for not knowing how to use latex)

from x=0 to x=1 of:

∫cos([(2n+1)(pi)/2]x)*cos([(2m+1)(pi)/2)]x) dx = 0, for m ≠ n

2. Relevant equations
The question tells me to use integral tables.

3. The attempt at a solution
Using integral tables, I got

∫(1/2) [cos((A-B)x) + cos((A+B)x)] dx from 0 to 1.

A, B are equal to (2n+1)pi/2 and (2m+1)pi/2 respectively.

However, when I evaluate the integral and I get sines, the x merely becomes a 1 and I get a little confused after that.

Any help would be greatly appreciated.

Last edited: Oct 7, 2012
2. Oct 7, 2012

### voko

What is $\int_0^1 \cos \alpha x dx$?

3. Oct 7, 2012

### royblaze

I believe it is sin(ax)/a evaluated from 0 to 1, so that would be sin(a)/a, correct?

4. Oct 7, 2012

### voko

Exactly correct. Now substitute A + B and A - B.

5. Oct 7, 2012

### royblaze

For the integration from 0 to 1, for (a+b) and (a-b) respectively I get:

sin(a+b)/(a+b)

sin(a-b)/(a-b)

Since it is basically a sum of integrals, I can just sum these (with the accompanying 1/2), correct? But I don't know what else after that: this is where I got confused.

6. Oct 7, 2012

### voko

There is one problem with you derivation. When A = B, you cannot integrate cos (A - B)x as sin (A - B)x / (A - B), because A - B = 0. However, cos (A - B)x = cos 0 = 1, and that can be integrated. What is the result? Note this is for m = n.

In all the other cases, what are sin (A + B) and sin (A - B)?

7. Oct 7, 2012

### royblaze

Right, okay, so if m = n, then the cosine term immediately becomes cos(0) = 1, and the integral of 1 in this case would just be the variable x. But why are we considering m = n?

As for the second question... I guess that A+B would be a positive number and A-B would be a negative number??? In all other cases... I'm a little confused by what you mean. Hopefully I am interpreting correctly.

8. Oct 7, 2012

### voko

We are considering m = n because A - B = 0 in the integrand is not a cosine but a constant.

For other cases, look at what A - B and A + B really are. What is the sine of them?

9. Oct 7, 2012

### royblaze

Ah! You mean the literal sine of the quantities A + B, A - B, right?

So the sum of A and B are sums of (σ∏)/2, where σ is some constant made by the sum of two odd numbers. And the sum of any odd numbers is an even number. Which puts the A + B into the form of just σ∏. Sine of a multiple of ∏=0.

The difference of A and B would give the same type of result, right?

I hope I'm following correctly. I still don't know the grounds for considering m = n when the question asks to prove the case for m ≠ n. I'm guessing this will all be revealed through this work.

10. Oct 7, 2012

### royblaze

I thought I got the proof, but I did not...

Last edited: Oct 7, 2012
11. Oct 7, 2012

### royblaze

Oh what the hell I'll just say it.

Since any sum of A and B (negative sum or otherwise) gives an even number multiple of ∏, then the sin (A+B) and sin (A-B) terms would give zero.

12. Oct 7, 2012

### voko

Exactly. Unlike the m ≠ n case.

13. Oct 8, 2012

### royblaze

In the m ≠ n case though, it is just a sum of ∏/2 terms. And any sum (negative or otherwise) of ∏/2 terms gives a term of ∏, which for sin = 0. So if n = 1000 or m = 12312412, the sine of A + B or A - B would still be zero, no?

14. Oct 8, 2012

### royblaze

By turning the product of the cosines into a sum of cosines, then it becomes clear that the integral in question would be zero after evaluation, I believe.

15. Oct 8, 2012

### voko

You have proven this, but since you sound a bit unsure, I will repeat that for you :)

Both A and B are an odd number of pi/2, so both their sum and their difference is an even number of pi/2, that is to say, some integer number of pi, the sine of which is zero. Hence the integral is zero unless m = n, when there is a constant term, which does not integrate to a sine.

16. Oct 8, 2012

### royblaze

Awesome Voko, thanks so much. You did an excellent job of leading me to my own conclusion. Thanks again!

It would be nice if I could give you a +1 or something along those lines. Great help!