# Thread: Prove these functions are orthogonal

1. ## Prove these functions are orthogonal

Hi, can anyone help me solve this:

2 functions f and g are orthogonal on [a,b] if:

$\displaystyle \int_a^b f(x)g(x)dx = 0$

Prove that $\displaystyle f_{m}(x) =$sin$\displaystyle mx$ and $\displaystyle f_{n}(x) =$sin$\displaystyle nx$ are orthogonal on the interval $\displaystyle [-\pi,\pi]$ if m and n are integers such that $\displaystyle m^2$ ≠ $\displaystyle n^2$

Thanks.

2. Originally Posted by coldfire
Hi, can anyone help me solve this:

2 functions f and g are orthogonal on [a,b] if:

$\displaystyle \int_a^b f(x)g(x)dx = 0$

Prove that $\displaystyle f_{m}(x) =$sin$\displaystyle mx$ and $\displaystyle f_{n}(x) =$sin$\displaystyle nx$ are orthogonal on the interval $\displaystyle [-\pi,\pi]$ if m and n are integers such that $\displaystyle m^2$ ≠ $\displaystyle n^2$

Thanks.
Please post what you've done and state where you get stuck.

Note that $\displaystyle \sin (nx) \sin (mx) = \frac{1}{2} \left(\cos ([n - m]x) - \cos ([n + m]x) \right)$.

3. Originally Posted by coldfire
Hi, can anyone help me solve this:

2 functions f and g are orthogonal on [a,b] if:

$\displaystyle \int_a^b f(x)g(x)dx = 0$

Prove that $\displaystyle f_{m}(x) =$sin$\displaystyle mx$ and $\displaystyle f_{n}(x) =$sin$\displaystyle nx$ are orthogonal on the interval $\displaystyle [-\pi,\pi]$ if m and n are integers such that $\displaystyle m^2$ ≠ $\displaystyle n^2$

Thanks.
Using the identitiy that

$\displaystyle \sin(u) \sin(v)=\frac{1}{2}(\cos(u-v)-\cos(u+v))$

So in your case you get

$\displaystyle \sin(mx) \sin(nx)=\frac{1}{2}(\cos((m-n)x)-\cos((m+n)x))$

It should all be down hill from here.

EDIT: haha too slow