Extrema of function $y=e^{-kx} \sin(nx)$ form a geometric progression I found this problem in Sinhalese and I'm trying my best to translate it:

Prove that the graph of $y=e^{-kx}\sin(nx)$ has a series of minimum and maximum ordinates, along the $x$-axis at every integer $\pi/n$ interval. And prove that they make a decreasing sequence according to a geometric progression for all $k\in\mathbb{R}^+$.

I took the derivative and found its roots. Hence, found there are minimums and maximums at every integer $\pi/n$ interval. I did not go further because I was not really sure what the problem asked.
Please explain how to prove it.
Thanks.
 A: The function vansihes at the points $m\pi/n$ hence it admits alternatively the minimum and the maximum on the intervals $ [(m-1)/n,m\pi/n],$ depending on the parity of $m.$ The equation for the extremal points is $$ -ke^{-kx}\sin(nx)+ne^{-kx}\cos(nx)=0$$ hence $\tan(nx)={n\over k}.$ We obtain infintely many solutions $$x_m={m\pi\over n}+{\arctan(n/k)\over n}$$ Let $\alpha = {\arctan(n/k)\over n}$ We get $$f(x_m)=(-1)^me^{-km\pi/n}e^{k\alpha}$$ The maxima, i.e. $f(x_{2m})$ form a decreasing geometric progression.
A: [I just discovered that a completely similar question (with $\cos$ instead of $\sin$) had already been asked on MathSE here].

An initial remark: the problem is correctly set if one interprets "in every $\pi/n$ interval" as "in every interval $[k \pi/n, (k+1) \pi/n]$"
First solution:
Let $$f(x):=e^{-kx} \sin(nx)\tag{1}$$
Have a look at the following figure, made with Geogebra, featuring the different extrema $(x_p,y_p), \ \ p=0,1,2...$ and allowing the computation of the ordinates $y_p$.

Let us find the abscissas $x_p$ of the extremas by solving equation $f'(x)=0$:
$$e^{-kx}(-k \sin(nx) + n \cos(nx))=0$$
giving :
$$\underbrace{\tfrac{\sin(nx)}{\cos(nx)}}_{\tan(nx)}=\frac{n}{k}\tag{2}$$
Setting $a=\tan^{-1}\left(\frac{n}{k}\right) \ \iff \ \tan(a)=\frac{n}{k},$
(2) is equivalent to $\tan(nx)=\tan(a)$, itself equivalent to
$$nx=a+p \pi \ \ \text{for a certain integer} \ p \ge 0$$
because $\tan$ is a $\pi$-periodic function. Finally
$$x=x_p=\frac{a}{n}+p \frac{\pi}{n}, \ \text{for} \ p=0,1,2 \cdots \tag{3}$$
Remark: please note that the abscissas of the extremas form an arithmetic sequence.
Numerical testing (using criteria $y_{p-1}y_{p+1}=y_p^2$) on the first values of $p$ have convinced me that sequence $y_p=f(x_p)$ is indeed a geometric progression with a negative ratio $-1<r<0$.
Here is the proof:
$$f(x_p)=e^{-k\left(\frac{a}{n}+p \frac{\pi}{n}\right)}\sin \left(n \left(\frac{a}{n}+p \frac{\pi}{n}\right)\right)$$
$$y_p=e^{-k\left(\frac{a}{n}+p \frac{\pi}{n}\right)}\sin(a+p \pi) $$
$$y_p=e^{-\frac{ka}{n}}(e^{-\frac{k \pi}{n}})^p (-1)^p\sin(a) $$
$$y_p=\underbrace{e^{-\frac{ka}{n}}\sin(a)}_{\text{first term} \ y_0}(\underbrace{-e^{-\frac{k \pi}{n}}}_{\text{ratio} \ r})^p  \tag{4}$$
proving that sequence $y_p$ is a geometric progression with negative ratio.
Remark: in fact, nothing prevents from taking negative values of $p$ : otherwise said, the maxima/minima for negative values of the variable could have been considered as well.
Second solution: (sketched)
$e^{-kx} \sin(nx)$ is the imaginary part of $e^{-kx} e^{inx}$. Otherwise said, setting $nx=\theta$, our issue amounts to find the ordinates of the maxima/minima of logarithmic spiral with polar equation :
$$\rho=e^{-\tfrac{k}{n} \theta}\tag{5}$$
which clearly happen when
$$\theta=\theta_p=\frac{\pi}{2}+p \pi \tag{6}$$
(if $p$ is even, it is a maxima, otherwise a minima)
Replacing into (5) the expression of $\theta$ given by (6), we get
$$\underbrace{\rho_p}_{y_p}=e^{-\tfrac{k}{n}(\tfrac{\pi}{2}+p \pi)}=e^{-\tfrac{k \pi}{2n}}\left(e^{-\tfrac{k \pi}{n}}\right)^p$$
which differs from (4) because the sign information is - naturally - not present in this polar equation.
Therefore, (5) appears as a kind of continuous version of (4).
I don't expand furthermore this solution because it necessitates to know some techniques on polar equations that are not necessarily commonplace.
