# Prove sin(mx)cosn(nx) is odd

1. Oct 3, 2012

### tainted

1. The problem statement, all variables and given/known data
Prove the following formula
$\int_{-\pi}^{\pi} \sin(mx)\cos(nx)\,dx = 0\\ (m, n = \pm 1, \pm 2, \pm 3, ...)$
2. Relevant equations
$\sin(A)\cos(B) = \frac{1}{2}[\sin(A-B)+\sin(A+B)]$

3. The attempt at a solution

$\int_{-\pi}^{\pi} \sin(mx)\cos(nx)\,dx\\ \int_{-\pi}^{\pi} \sin(mx-nx)+\sin(mx+nx)\,dx\\ \int_{-\pi}^{\pi} \sin((m-n)x)+\sin((m+n)x)\,dx$
Edit
$\int_{-\pi}^{\pi} \sin((m-n)x)\,dx + \int_{-\pi}^{\pi} \sin((m+n)x)\,dx$
the function, sine of any constant times x, is an odd function, therefore when it is integraded from -a to a, its value is 0.
Therefore, we get
$\int_{-\pi}^{\pi} \sin((m-n)x)\,dx + \int_{-\pi}^{\pi} \sin((m+n)x)\,dx = 0 + 0 = 0$
Would that be sufficient in the proof?

Thanks guys!

Last edited: Oct 3, 2012
2. Oct 3, 2012

### vela

Staff Emeritus
Just do the integrals, taking care to treat the case where m=n separately. You know how to integrate sin kx, right?

3. Oct 3, 2012

### LCKurtz

What do you know about odd and even functions and integrals? Are sines and cosines odd or even? Can you draw any conclusions about those integrands?

4. Oct 3, 2012

### tainted

Ok well of course -sin(x) = sin(-x) but does that hold true when there is a constant in front of x?

5. Oct 3, 2012

### LCKurtz

x can be anything in that identity, can't it? Like, for example, cx.

6. Oct 3, 2012

### HallsofIvy

Staff Emeritus
a(-x)= -(ax).

7. Oct 3, 2012

### tainted

Ok thanks guys! I think I got it, could you tell me if my attempted answer to the question appears to be sufficient?

8. Oct 3, 2012

### vela

Staff Emeritus
Yeah, that's sufficient. But can you see how could have done the whole problem in one step?