The number of zeros of function $f(x)=\sin(x)-m\sin(kx)-n\sin(x-kx)\text{ on }(0,\pi)$, where $m,n>0$ and $0<k<1$

Prove the former function has $0$ or $1$ zero on $(0,\pi)$.

I came up with this question because I want to think through one of my previous question: Prove ABC,A'B'C' are congruent:D is on BC,D' is on B'C', $\angle BAD \angle CAD= \angle B'A'D' \angle C'A'D', AB=A'B', AC=A'C', AD=A'D'$, working for a formal proof.

I'm sorry to ask another question here, but recently I've been working hard in this question for days, and none of the ways works. And I think it might worth opening another question for the proof of 0/1 zero(s).

First I tried to prove the monotonicity, so I calculated the difference of the function: $$f'(x)=\cos(x)-mk\cos(kx)-n(1-k)\cos(x-kx).$$ However, it is not always positive or negative, and what's even worse, the difference is non-monotonic!

It seems hard, and I wish someone can give me a hand.

  • $\begingroup$ There is a zero, right at the origin: $f(0) = \sin{(0)} - m*sin{(k(0))} - n * sin{(0 - k(0))}$, which results in $f(0) = 0$, $\therefore$ the function has a root at $(0, 0)$. $\endgroup$ Aug 8, 2021 at 14:31
  • $\begingroup$ $(1-x)$ seems like a typo. Should it be $(1-k)$? $\endgroup$
    – user58697
    Aug 8, 2021 at 20:55
  • $\begingroup$ @SnipingPoodle No, (0,2pi) does not contain 0. $\endgroup$ Aug 9, 2021 at 2:10
  • $\begingroup$ @user58697 Yes, I've just fixed that typo. $\endgroup$ Aug 9, 2021 at 2:11


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