# Proof of an inequality for a linear combination of three trigonometric functions

Given a function $$f(t) = k_{1} \sin(t+\alpha) + k_{3} \sin(3t+\beta) + k_{5} \sin(5t+\gamma)$$ where $k_{1}, k_{3}$ and $k_{5}$ are all positive parameters, and the three phase angles, $0<\alpha<\pi/2, 0<\beta,\gamma<2\pi$. The problem is: if $f(t)$ has only two extreme points in a period $[0,2\pi]$, say $t_1$, $t_2 \in [0,2\pi]$, due to anti-symmetry of $f(t)$, without loss of generality，$t_2$ = $t_1 + \pi$， then does the following inequality always hold for $k_{1}, k_{3}$ and $k_{5}$:

$$k_{1}> k_{3} > k_{5}$$

My first impression was to use time derivative of $f(t)$ and root of $f(t)$, i.e.,

$$f'(t_1) = k_{1}\cos(t_1+\alpha) + 3k_{3} \cos(3t_1+\beta) + 5k_{5} \cos(5t_1+\gamma)=0$$ $$f(t_1+\pi/2) = k_{1}\cos(t_1+\alpha) -k_{3}\cos(3t_1+\beta) + k_{5} \cos(5t_1+\gamma)=0$$ but I have no idea how to go further. Maybe I need some additional constraints for the phase angle parameter. I need some help here, please.

P.S. I've seen that the pattern of such relation $$k_{1}> k_{3} > k_{5}$$ is true in nearly all cases that I've encountered, but I have never seen any theoretical work as to why this is the case.

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Probably not - if $k_3$ and $k_5$ are very small there will be only two extreme points, whether $k_3>k_5$ or the other way round. –  Carl Nov 12 '13 at 22:59
I am not sure if you are right since it is not easy to find an specific anti-example. To make the problem simpler, $f(t)=k 1 sin(t+α)+k 3 sin(3t+β)$, with the same conditions, does the following inequality hold for $k_1$, $k_3$:$$k_1 >k_3$$? –  Alok Samanta Nov 13 '13 at 11:23
In the simpler case, I think yes: Find the derivative, then the $3k_3$ term will dominate the $k_1$ term, leading to six critical points in $[0,2\pi)$. For the original case, I think $\sin(t)+0.01\sin(3t)+0.02\sin(5t)$ has only two critical points. –  Carl Nov 13 '13 at 13:06
@Carl: Thanks. You are right that the original case might be not true. I just checked the example you mentioned and the numerics show only two critical points. But for the simple case, only $k_1, k_2$, it must be true. In addition, do you agree that $$f(t_1+π/2)=0$$ ? –  Alok Samanta Nov 19 '13 at 13:15
Yes, my example clearly has critical points at $t=(n+1/2)\pi$ and zeros at $t=n\pi$ for all integers $n$. –  Carl Nov 19 '13 at 17:21