Proving the Oddness of sin(mx)cos(nx) using Integration

[tainted]

Homework Statement

Prove the following formula
$\int_{-\pi}^{\pi} \sin(mx)\cos(nx)\,dx = 0\\ (m, n = \pm 1, \pm 2, \pm 3, ...)$

Homework Equations

$\sin(A)\cos(B) = \frac{1}{2}[\sin(A-B)+\sin(A+B)]$

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!

 

[vela]

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

 

[LCKurtz]

Homework Statement

Prove the following formula
$\int_{-\pi}^{\pi} \sin(mx)\cos(nx)\,dx = 0\\ (m, n = \pm 1, \pm 2, \pm 3, ...)$

Homework Equations

$\sin(A)\cos(B) = \frac{1}{2}[\sin(A-B)+\sin(A+B)]$

The Attempt at a Solution

$\int_{-\pi}^{\pi} \sin(mx)\cos(nx)\,dx\\ \int_{-\pi}^{\pi} \sin(mx-nx)+\sin(mx+nx)\,dx\\ \int_{-\pi}^{\pi} \sin((m-n)x)+\sin((m+n)x)\,dx\\$

Now I would imagine that there is some identity to either evaluate these or to prove that these are odd.
!

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?

 

[tainted]

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

 

[LCKurtz]

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

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

 

[HallsofIvy]

a(-x)= -(ax).

 

[tainted]

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

 

[vela]

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