I apologize for the lack of context. I was given the following information that $$\frac{dP}{dt} = kP \cdot \cos(kt),$$ where $P(t)$ is the size of population at $t$ hours and $k$ is some constant. I would like to determine the ratio of the minimum size of the population to the maximum population size.
Using separation of variables, I eventually got to $$P(t) = Ae^{\sin(kt)},$$ where $A = e^C$ for some constant $C.$
I suppose the minimum value sin can be is $-1,$ and the maximum is $1.$ Then, the ratio is $\frac{Ae^{-1}}{Ae},$ and this is equal to $\frac{1}{e^2}.$
Is this reasoning/logic correct?
Ultimately, you are correct, but there is something slightly deeper at play here.
Lemma. Given an everywhere differentiable function $f(x),$ the composition $g(x) = e^{f(x)}$ has a local maximum (resp. minimum) at $x = a$ if and only if $f(x)$ has a local maximum (resp. minimum) at $x = a.$
Proof. By the Chain Rule, we have that $g'(x) = e^{f(x)} f'(x).$ Considering that $e^{f(x)} > 0$ for all $x,$ we have that $g'(x) = 0$ if and only if $f'(x) = 0.$ Consequently, the critical points of $f(x)$ and $g(x)$ are the same. Even more, the sign of $g'(x)$ is the same as the sign of $f'(x),$ hence the claim holds.
Our case is $P(t) = Ae^{\sin(kt)}.$ Considering that $\sin(kt)$ is an everywhere differentiable function, in order to find the local extrema of $P(t),$ it suffices (by the lemma) to find the local extrema of $\sin(kt).$ Of course, these occur periodically when $kt = \frac{(2n + 1) \pi}{2}$ for each integer $n,$ so the maximum (resp. minimum) value of $\sin(kt)$ is $1$ (resp. $-1$). Consequently, by the lemma, the maximum value of $P(t)$ is $Ae,$ and the minimum value of $P(t)$ is $Ae^{-1}.$