I couldn't find this anywhere here, so I'll ask it.
How would I show: $$\{\;\sin((2n+1)x)\;\}$$ on $[0,\pi/2]$, where $n>0$ and is an integer, is an orthogonal set with respect to the inner product $\int_0^{\pi/2}f(x) g(x) dx$?
I've tried a couple different ways, and they all explode into crazy integrals. Is there a clean way to show this?
On
Observe that for odd $\;k,\,m\;$ , we have (using trigonometric identities for products):
$$\langle \sin kx,\,\sin mx\rangle=\int_0^{\pi/2}\sin kx\sin mx\,\mathrm dx=\frac12\int_0^{\pi/2}\left(\cos(k-m)x-\cos(k+m)x\right)\,dx=$$
$$=\begin{cases}\frac12\int\limits_0^{\pi/2}\left(1-\cos2kx\right)\,dx=\frac12\left(\frac\pi2\right)-\left.\frac1{2k}\sin2kx\right|_0^{\pi/2}=\frac\pi4\,,&k=m\\{}\\ \frac12\left.\left(\frac1{k-m}\sin(k-m)x-\frac1{k+m}\sin(k+m)x\right)\right|_0^{\pi/2}=0\,,&k\neq m\end{cases}$$
Thus, multiplying each element of that set by $\;\cfrac4\pi\;$ you get in fact an orthonormal basis.
Although it might not seem to be the case, the integrals you presumably wrote down actually can be evaluated by elementary (if clever) means. This note illustrates two ways of doing so: one using trigonometric identities to change the product $\sin(2n+1)x \cdot \sin(2m+1)x$ into a more easily integrated form, and a second using repeated integration by parts.