I was watching the MIT - Lec 15 on Fourier Series (Linked Below ) and the professor

mentioned a proof for orthogonal functions using trig identities.
Does anyone know that proof?

The other two proofs, in case you were wondering,
where using a second order ODE and complex exponentials.


http://www.youtube.com/watch?v=EWWw0jryj…

Calculation that the functions {sin(mx) , cos(nx) : integers m > 0, n >=0 } are
orthogonal on [-pi,pi] :

Use the identities sin(a + b) = sin(a) cos(b) + cos(a) sin(b)

and cos(a + b) = cos(a) cos(b) -- sin(a) sin(b)

(1) 2 sin(mx) cos(nx) = sin((m+n)x) + sin((m-n)x)


Note that for any non-0 integer k,

Int_(-pi.pi) sin(kx) dx = -- 1/k [cos(k pi) - cos(-kpi)] = 0

If k = 0 it's trivial that the integral = 0.

(Likewise Int(-pi,pi) cos(kx) dx = 0 unless k = 0)

Then by identity (1), Int_(-pi,pi) sin(mx) cos(nx) dx = 0

(2) 2 sin(mx) sin(nx) = cos((m+n)x) -- cos ((m-n)x)

If m & n are unequal positive integers, it follows that

Int_(-pi,pi) sin(mx) sin(nx) dx = 0
------

I leave the final case to you:

Int_(-pi,pi) cos(mx) cos(nx) dx = 0 if m & n are unequal integers >= 0

You just use the product to sum identities.