TITLE: Proving that a set is an orthonormal basis
QUESTION [1 upvotes]: Any ideas on how to quickly show that
$$
\left( \frac{1}{\sqrt{2\pi}}, \frac{\sin(x)}{\sqrt{\pi}}, \frac{\sin(2x)}{\sqrt{\pi}}, ..., \frac{\sin(nx)}{\sqrt{\pi}}, \frac{\cos(x)}{\sqrt{\pi}}, \frac{\cos(2x)}{\sqrt{\pi}}, ..., \frac{\cos(nx)}{\sqrt{\pi}} \right)
$$
is an orthonormal basis of vectors in $C[-\pi, \pi]$ (the vector space of continuos real valued functions on $[-\pi, \pi]$) where the inner product is defined as
$$
\left<f, g\right> = \int_{-\pi}^{\pi} f(x)g(x) \, dx
$$
I found some integration formulae that allow you to show it but I don't really want to have to re-derive all of them on my homework. I'm wondering if there's a simpler argument that you can make.

REPLY [2 votes]: To handle the product of trig functions with $n\ne m$, you can integrate by parts twice, ending up with the same integral multiplied by $n^2/m^2$. Boundary terms disappear because of periodicity.  
The above is a more conceptual proof than messing with product-sum formulas, because it uses what is really   essential here: the basis is formed by eigenfunctions of a symmetric operator.  
The above does not work for $\cos nx \sin nx$, and for a good reason: two eigenfunctions with the same eigenvalue have no need to be orthogonal. These two happen to be orthogonal due to  our convenient definition  of trigonometric functions. Simple reason: the product is odd, and the interval of integration is symmetric.
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