Why we need to mention the scalar product of $\cos (nx), \sin (nx)$?

Question

I found the following text -

The functions $\cos (nx), n = 0, 1, 2, \cdots$ and $\sin (nx), n = 1, 2, \cdots $ which are known to be orthogonal with respect to the standard scalar product on $(-\pi, \pi)$.

The source of the problem is -

Question: Plz explain why we author needed to mention "..the functions $\cos (nx), n = 0, 1, 2, \cdots$ and $\sin (nx), n = 1, 2, \cdots $ which are known to be orthogonal with respect to the standard scalar product on $(-\pi, \pi)$".... as we dont need the scalar product of $\cos (nx), \sin (nx)$ in the above example?

Note:

We are doing scalar product of $f$ and $\sin nx$, scalar product of $f$ and $\cos nx$, scalar product of $\sin nx$ and $\sin nx$, scalar product of $\cos nx$ and $\cos nx$, but we never used scalar product of $\cos nx$ and $\sin nx$, then why it was mentioned? what is the meaning of that part of the sentence? Thanks.

Answer 1

Given that

$$f\left(x\right) = a_0 +\sum _{n=1}^\infty \left(a_n\cos nx+b_n\sin nx\right)$$ We get the expression for $b_n$ as follows $$\left\langle f\vert \sin mx \right\rangle = \sum _{n=1}^\infty \left\langle \left(a_n\cos nx+b_n\sin nx\right)\vert \sin mx \right\rangle$$ $$=b_m\left\langle \sin mx\vert \sin mx \right\rangle$$ Using both that $\left\langle \sin mx\vert \sin nx \right\rangle=0$ for $n\neq m$ and that $\left\langle \sin mx\vert \cos nx \right\rangle=0$, i.e. that sin and cos are orthogonal.

Answer 2

The scalar product is used right after, when the author writes$$b_n=\frac{\langle\sin nx|f\rangle}{\langle\sin nx|\sin nx\rangle}.$$

Answer 3

We do in fact need the scalar product of $\sin(nx)$ and $\cos(nx)$. Suppose we want to take a Fourier series of $\sin(mx)$ ($m\in \mathbb{N}$). We know, based on the definition, that $a_n=0 ~\forall n$, and $b_n=1$ if $n=m$ or $0$ otherwise. Thus, $$a_n=\int_{-\pi}^{\pi}\sin(mx)\cos(nx)\mathrm{d}x$$ In order for consistency, this needs to be $=0$ $\forall n$, or in other words, we need $\sin(mx)$ and $\cos(nx)$ to be orthogonal.