Orthogonal complement of span$\langle\, \sin, \cos\,\rangle$ in $L^2$

Question

Let $V= \{ f(x) = \alpha \sin x + \beta \cos x : \, \alpha, \beta \in \mathbb{R} \} \subset L^2[0,\pi] $.
I am asked to find $V^{\bot}$.

I recognize that $V$ is a closed subspace of $L^2[0,\pi]$ (since it is a finite-dimensional subspace of a normed vector space).
So (since $H=L^2[0,\pi] $ is a Hilbert space) there holds: $H = V \oplus V^{\bot}$ and $V^{\bot} = \operatorname{Im}\big( I - P_{V} \big)$. But I cannot go deeper.

Is it possible to find a more precise identification for $V^{\bot}$ in your opinion?

Answer 1

By Fourier Theory, $\{\frac{1}{\sqrt{\pi}}e^{inx/2}\}_{n\in \mathbb{Z}}$ is an orthonormal basis of $L^2([0,\pi])$. If you don't know it, you can see that their span is uniformly dense in the continuous functions with Stone-Weierstrass and then, you can apply Urysohn's Lemma to see that the continuous functions are dense in $L^2([0,\pi])$. Note that we've used that $[0,\pi]$ is a finite measure space. Anyway, you're presumably expected to know this result one way or another.

Once you have this, you can use Euler's identity to check that $\textrm{span}\{\sin,\cos\}=\textrm{span}\{e^{ix},e^{-ix}\}$ and hence, $$ V^{\perp}=\overline{\textrm{span}\{e^{i nx/2}\}_{n\in \mathbb{Z}\setminus\{-2,2\}}} $$

Answer 2

$V^\perp$ consists of all functions written in terms of the inner product $\langle f,g\rangle=\int_0^{\pi}f(t)g(t)dt$: $$ f-\frac{\langle f,\sin(x)\rangle}{\langle \sin(x),\sin(x)\rangle}\sin(x)-\frac{\langle f,\cos(x)\rangle}{\langle\cos(x),\cos(x)\rangle}\cos(x),\;\;\; f\in L^2[0,\pi]. $$ This is because $\langle \sin(x),\cos(x)\rangle =0$.