How to derive $\frac1\pi \int_{-\pi}^{\pi}f(t)\sin nt \;\mathrm{d}t$ from $\frac{\langle\sin nx|f\rangle}{\langle \sin nx|\sin nx\rangle}$?

Question

How to get $$\frac{\langle\sin nx|f\rangle}{\langle \sin nx|\sin nx\rangle}=\frac1\pi \int_{-\pi}^{\pi}f(t)\sin nt \;\mathrm{d}t?$$ To be specific, $\langle \sin nx|\sin nx\rangle$ becomes $\frac1\pi?$

I have attached the excerpt below -

Answer 1

I think you meant the following:$$\pi\stackrel{?}{=}\langle\sin nx | \sin nx \rangle:=\int_{-\pi}^{\pi} (\sin nx)^*\cdot\sin nx\:dx=\int_{-\pi}^\pi\sin^2 nx\:dx$$ and indeed, the integral does equal $\pi$ (can be easily seen by using $\cos2nx=1-2\sin^2 nx$).

EDIT (OPs request) We have to evaluate $$ \int_{-\pi}^{\pi} \sin^2 nx\:dx=\frac{1}{2}\int_{-\pi}^\pi(1-\cos 2nx)\:dx=\frac{1}{2}\left.\left(x-\frac{\sin 2nx}{2n}\right)\right|_{-\pi}^{\pi}=\pi$$