I want to prove that $S = \left\{\sqrt{\tfrac{2}{\pi}}\sin(kx):k\in\mathbb{N}\right\}$ forms an orthonormal basis of $L^2[0,\pi]$. I may use the fact, that $B = \left\{\sqrt{\tfrac{2}{\pi}}\sin(2kx):k\in\mathbb{N}\right\} \cup \left\{\sqrt{\tfrac{2}{\pi}}\cos(2kx):k\in\mathbb{N}\right\} \cup \{ \tfrac 1 \pi \}$ is an orthonormal basis for $L^2[0,\pi]$.
To show, that $S$ is an orthonormal system is easy. But I have problems to show, that the span of $S$ is dense in $L^2[0,\pi]$.
I could already show that $\left\|\tfrac1\pi\right\|^2 = \sum_{k=1}^\infty \left|\langle\tfrac 1\pi,\sqrt{\tfrac{2}{\pi}}\sin(kx)\rangle\right|^2$. Does this imply that $\tfrac1\pi$ is in the closure of the span of $S$?
So am I right that the only think left to show for me is $\left\|\cos(2mx)\right\|^2 = \sum_{k=1}^\infty \left|\langle\cos(2mx),\sqrt{\tfrac{2}{\pi}}\sin(kx)\rangle\right|^2$? (which I couldn't do so far) My idea is to prove that $B$ is in the closure of the span of S and thus the span of $S$ dense in $L^2[0,\pi]$ (because the span of $B$ is dense in $L^2[0,\pi]$).
Yes, $$\begin{equation}\lVert f\rVert^2 = \sum_{k=1}^\infty \Bigl\lvert\langle f,\, \sqrt{\frac{2}{\pi}}\sin (kx)\rangle\Bigr\rvert^2\end{equation}$$ proves that $f$ is in the closure of the span of $S$. You can write $f = f_S + (f - f_S)$, where $f_S$ is the orthogonal projection onto the closure of the span of $S$. Then
$$\lVert f\rVert^2 = \lVert f_S\rVert^2 + \lVert f - f_S\rVert^2,\text{ and } f_S = \sum_{k = 1}^\infty \langle f,\, \sqrt{\frac{2}{\pi}}\sin (kx)\rangle\cdot \sqrt{\frac{2}{\pi}}\sin (kx)$$
(modulo possibly conjugating/swapping the order of the factors in the inner product if you consider complex-valued functions). So if you have the first equation, you know that $\lVert f - f_S \rVert = 0$, or $f = f_S$.