I want to prove it's a basis respect the inner product $\langle f,g\rangle=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \overline{g(x)}dx$.
I have to calculate and prove :
$\langle\frac{1}{\sqrt 2},\frac{1}{\sqrt 2}\rangle=\langle\cos(kx),\cos(kx)\rangle=\langle\sin(kx),\sin(kx)\rangle=1$
$\langle\frac{1}{\sqrt 2},\cos(kx)\rangle=\langle\frac{1}{\sqrt 2},\sin(kx)\rangle=0$
$\langle f,g\rangle=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \overline{g(x)}dx$ so can I use Euler's formula?
You have 8 intergrals to solve, you should have learned the techniques to solve them in calc 1.
$\frac 1{\pi}\int_{-\pi}^{\pi} (\frac {1}{\sqrt 2})(\frac {1}{\sqrt 2}) \ dx = 1$
$\forall k\in \mathbb N, \frac 1{\pi}\int_{-\pi}^{\pi} \cos^2 kx \ dx = 1$
$\forall k\in \mathbb N, \frac 1{\pi}\int_{-\pi}^{\pi} \sin^2 kx \ dx = 1$
$\forall k\in \mathbb N, \frac 1{\pi}\int_{-\pi}^{\pi} \frac {1}{\sqrt 2}\cos kx \ dx = 0$
$\forall k\in \mathbb N, \frac 1{\pi}\int_{-\pi}^{\pi} \frac {1}{\sqrt 2}\sin kx \ dx = 0$
$\forall k,l\in \mathbb N | k\ne l, \frac 1{\pi}\int_{-\pi}^{\pi} \cos kx\cos lx \ dx = 0$
$\forall k,l\in \mathbb N | k\ne l, \frac 1{\pi}\int_{-\pi}^{\pi} \cos kx\sin lx \ dx = 0$
$\forall k,l\in \mathbb N | k\ne l, \frac 1{\pi}\int_{-\pi}^{\pi} \sin kx\sin lx \ dx = 0$
The first 5 shold be pretty simple, the last 3 use a variant of $\cos A\cos B = \frac 12 (\cos (A+B) + \cos (A-B))$
