Question: Let $f \in L^2([0, 2\pi])$. Show that there exists unique $\alpha, \beta \in \mathbb{R}$ so that $$\int_0^{2\pi}|f(x) - \alpha( \sin x + \cos x) - \beta \sin 3x|^2 \, dx \leq \int_0^{2\pi}|f(x) - a( \sin x + \cos x) - b\sin 3x|^2 \, dx\ \text{for all}\ a,b \in \mathbb{R}.$$
My attempt: This feels like an application of Bessel inequality, using the orthonormal system $\{a_n \cos n x, b_n \sin n x\}$. However, the inner-product in the Hilbert space $L^2([0, 2 \pi])$ involves taking product whereas the question expression involves difference of terms. Is there any way I work around it to apply Bessel's inequality?
Any hints are appreciated.
(On hindsight, I realised this does not involve Bessel's inequality, will be editing the title of this post) My attempt following the hints provided in the comments:
Let $u = \sin x + \cos x$ and $v = \sin 3x$. Then, \begin{align*} \langle f - \alpha u - \beta v, f - \alpha u - \beta v \rangle &= \alpha^2 \langle u,u \rangle - 2 \alpha \langle f ,u \rangle + \beta^2 \langle v, v \rangle - 2 \beta \langle f, v \rangle + \langle f,f \rangle\\ &= \left(\alpha \Vert u \Vert - \frac{\langle f ,u \rangle}{\Vert u \Vert} \right)^2 + \left( \beta \Vert v \Vert - \frac{\langle f ,v \rangle}{\Vert v \Vert} \right)^2 + \langle f, f \rangle - \left(\frac{\langle f ,u \rangle}{\Vert u \Vert} \right)^2 - \left( \frac{\langle f, v \rangle}{\Vert v \Vert} \right)^2 \end{align*} Hence, the minimum is attain when $\alpha = \frac{\langle f, u \rangle}{\Vert u \Vert^2}$ and $\beta = \frac{\langle f, v \rangle}{\Vert v \Vert^2}$.