How to show that $\displaystyle\frac{1}{\pi} \int_{-\pi}^{\pi} \cos(mx)\cos(nx) \,dx = \delta_{mn}$?
If you use $\cos(x)\cos(y)=\cos(x-y)+\cos(x-y)$, you get that
$\begin{align} \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(mx)\cos(nx) \,dx &= \frac{1}{\pi}\left( \int_{-\pi}^{\pi} \cos((m-n)x)\,dx+\int_{-\pi}^{\pi} \cos((m+n)x) \,dx \right) \\ &= \frac{1}{\pi}\left(\left[\frac{1}{m-n}\sin((m-n)x)\right]_{x=-\pi}^{x=\pi}+\left[\frac{1} {m+n}\sin((m+n)x)\right]_{x=-\pi}^{x=\pi}\right) \\ &= \frac{1}{\pi}\frac{2((n-m)\sin(\pi n+\pi m)+(n+m)\sin(\pi n-\pi m))}{2 n^2-2 m^2 }. \end{align}$
But now if I consider $m=n=k$, then integral goes to $0$ and if I consider $m \neq n$, then it still goes to $0$. $\delta_{mn}$ is Kronecker delta.