Orthogonality of 20220724 trigonometric function system

Trigonometric function system is defined as ：$$\{1,\sin x,\cos x,\sin 2x,\cos 2x,\cdots ,\sin nx,\cos nx,\cdots \}$$ Orthogonality means that the product of any two different functions is in the interval $[-\pi ,\pi ]$ The inner integral is 0, namely $${\int }_{-\pi }^{\pi }\sin nx\cos nx\text{ d}x=0$$$${\int }_{-\pi }^{\pi }\cos nx\cos mx\text{ d}x=0,m\ne n$$$${\int }_{-\pi }^{\pi }\sin mx\text{ d}x=0$$

For the first and third equations , because $\sin nx\cos nx$ and $\sin mx$ It's an odd function , So it was established .

For the second equation , The demonstration process is as follows ：
$$\begin{array}{rl}{\int }_{-\pi }^{\pi }\cos nx\cos mx\text{ d}x& =\frac{1}{m}{\int }_{-\pi }^{\pi }\cos nx\text{ d}\sin mx\\ & =\frac{1}{m}\cos nx\sin mx{|}_{-\pi }^{\pi }-\frac{1}{m}{\int }_{-\pi }^{\pi }\sin mx\text{ d}\cos nx\\ & =\frac{n}{m}{\int }_{-\pi }^{\pi }\sin mx\sin nx\text{ d}x\end{array}$$
It is further known that ,
$$\begin{array}{r}{\int }_{-\pi }^{\pi }\cos nx\cos mx\text{ d}x-{\int }_{-\pi }^{\pi }\sin mx\sin nx\text{ d}x=(1-\frac{m}{n}){\int }_{-\pi }^{\pi }\cos nx\cos mx\text{ d}x\end{array}$$
in addition $$\begin{array}{r}{\int }_{-\pi }^{\pi }\cos nx\cos mx\text{ d}x-{\int }_{-\pi }^{\pi }\sin mx\sin nx\text{ d}x={\int }_{-\pi }^{\pi }\cos ((m+n)x)\text{ d}x\end{array}=0$$
Then we can see , if $m\ne n$, Then there are $${\int }_{-\pi }^{\pi }\cos nx\cos mx\text{ d}x=0$$
another , if $m=n$, Obviously there is $${\int }_{-\pi }^{\pi }\cos nx\cos mx\text{ d}x>0$$.
Certificate completion .