Proving linear independence of the functions $\sin (nx)$ and $\sin (mx)$ without using Wronskian

I was asked to show the linear independence of the functions $\sin (nx)$ and $\sin (mx)$ where $m,n\in \mathbb N$, $m\neq n$. It can of course be done by computing the Wronskian, but I just wanted to avoid it and go old school. Here's a small sketch of my solution-

Let $f(x)=a\sin(mx)+b\sin(nx)=0\;\;\forall x$

We have three cases, namely both $m,n$ are even; exactly one of them is even; or both of them are odd.

Case I: Both are even

Let $m=2^{k_m}\times o_m$ and $n=2^{k_n}\times o_n$ where $o_m$ and $o_n$ are odd. WLOG, assume $k_m>k_n$. Put $x=\frac \pi{2^{k_m}}$ and note that $$f\left(\frac \pi{2^{k_m}}\right)=a\sin(\mathtt{odd}\times \pi)+b\sin\left(\mathtt{odd}\times \frac \pi{2^l}\right)$$ for some $l\in \mathbb N$.

So, $$f\left(\frac \pi{2^{k_m}}\right)=0+cb$$ for some $c\in [-1,1]\backslash\{0\}$, which completes the proof of $a=b=0$.

Case II: Exactly one of them is even

We will use the same idea as before. Let the highest power of $2$ in the even number be $k$ and use that to arrive at the form $$a\sin(\mathtt{odd}\times \pi)+b\sin\left(\mathtt{odd}\times \frac \pi{2^h}\right)$$ and complete the proof.

Case III: Both are odd

Evaluate $f$ at $x_1=\frac \pi 2$ and $x_2=\frac {3\pi}2$ to get the equations $$a+b=0\\ a-b=0$$ solving which gives us $a=b=0$.

Is this proof okay? Especially, is the last case okay? If not, how can we tackle that case? Also, is there a smarter choice of the $x$'s to arrive at the same conclusion?

One solution that I got is that just checking the zeroes of the two functions are enough to arrive at the conclusion of dependence. While that's a very nice observation, please note that my question is whether my proof (especially the third case about which I'm most worried) is okay, and whether there are smarter choice of $x$'s that can establish the same job.

Solution 1:

Your proof is correct, but long and tedious—- you must understand that if two non-zero ‘vectors’ $u$ and $v$ are linearly dependent, then there is a scalar $\lambda$ with $v=\lambda u$. (This is immediate from the definition of linear dependence you are using.) In particular if two functions are linearly dependent then they are essentially the same up to scaling. In particular they have the same periodicity and the same zeros.

But $\sin(mx)$ has zeros at $m\pi \mathbb{Z}=\{tm\pi | t\in \mathbb{Z}\}$. So if $\sin (mx)$ and $\sin (nx)$ are linearly dependent then the set of zeros is the same, which happens if and only if $m=\pm n$.

Your proof is essentially doing this, but splitting it into cases and checking explicit points instead of the entire collection of zeros.