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Let $S$ be a sum of cosines with the same amplitude and different $\omega$:

$$S(x) = A_0 \sum_{n = 1}^N \cos \left( \omega_n x + \phi_n \right)$$

that is $\omega_1 \neq \omega_2 \neq \ldots \neq \omega_n$.

What are the conditions to be verified for the $\omega_n$ and the $\phi_n$ in order to obtain, for at least one value of $x$, that $S(x) = NA_0$?

That is: when is it true that, given a number $N$ of summed cosines of amplitude $A_0$, the result will reach (at least once) the value $NA_0$?

For $N = 2$, a sufficient condition to be satisfied is: $\omega_1 = \omega_0 + \alpha$ and $\omega_2 = \omega_0 - \alpha$, along with $\phi_1 = \phi_0 + \beta$ and $\phi_2 = \phi_0 - \beta$. In that case, $S(x) = 2A_0 \cos \left( \omega_0 x - \phi_0 \right) \cos \left( \alpha x - \beta \right)$.

I would be pleased if who downvoted this question will leave a comment with a suggestion.

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    "it occurs when $x$ is such that both the cosines inside the sum have the same argument": no, this is not sufficient. – 2017-02-16
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    Are you looking for the maximum of $S$ or for the amplitude of the envelope? In the first case, you need $\omega_n x+\phi_n = 2k_n\pi,\;k_n\in\mathbb{Z}\forall n$ or $\omega_n x+\phi_n = (2k_n+1)\pi, \;k_n\in\mathbb{Z}\forall n$ to achieve $NA_0$. In the second case, $\omega_n x+\phi_n-(\omega_m x+\phi_m) = 2k_{nm}\pi$ is sufficient – 2017-02-16
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    @ReinhardMeier: how do you define the envelope ? – 2017-02-16
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    A description can be found at https://en.wikipedia.org/wiki/Envelope_(waves) – 2017-02-16
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    @ReinhardMeier: mh, how can you obtain an equation of the envelope and the maximum condition ? Can you elaborate ? – 2017-02-16
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    @ReinhardMeier I am looking for the amplitude of the envelope, but I supposed that it would coincide with the maximum of $S$. – 2017-02-16
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    I changed my question, after your right and legit observations. – 2017-02-16
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    @YvesDaoust To be honest, I only wanted to make sure that we are looking for the right thing. The statement with the "same cosine arguments" of the first version of the question indicated that the question aims at the maximum of the envelope. I do not know how this can be calculated in general, but I remember vaguely that the phases must match to get the maximum (maybe I am wrong). – 2017-02-16
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    @ReinhardMeier: if you look at a few samples, you soon see that envelopes are irrelevant, the curves are too chaotic. And matching the phases (if possible at all) is not sufficient to get a maximum. – 2017-02-16
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    That change of parameterization is of little use as the "condition" is always achieved. As before, it doesn't mean that $2A_0$ can be reached. – 2017-02-16

2 Answers 2

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If you are free to adjust the phases, then set $\phi_n=0$ for all $n$ and trivially

$$S(0)=NA_0.$$

If not, you can locate the extrema by canceling the first derivative. This gives the difficult equation

$$\sum_{n=0}^N\omega_n\sin(\omega_n x+\phi_n)=0.$$

This equation has infinitely many solutions which appear in a chaotic way when the $\omega$ are irrational mutiples of each other. I suspect that the sum never reaches $NA_0$ exactly (@Ingix proved it when $N=2$), but gets as close as you want for larger and larger $x$. But I can't prove this stronger property.

A case with $N=2$:

enter image description here

Don't believe that $2$ is ever exactly reached.

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Your proposition isn't even true for $N=2$. Set $\omega_1=1, \omega_2=2, \phi_1=\phi_2=\sqrt{2}$. In order for each cosine to be one, the argument must be an integer multiple of $2\pi$:

$$\begin{array}{c} \omega_1 x + \phi_1=2k_1\pi \\ \omega_2 x + \phi_2=2k_2\pi \end{array}$$

Subtracting the first equation from the second, one gets $ x = 2(k_2-k_1)\pi$ and putting the back into the first equation

$$ \sqrt{2}=\phi_1=2k_1\pi-2(k_2-k_1)\pi=(4k_1-2k_2)\pi$$

which would imply $\sqrt{2}$ to be an integer multiple of $\pi$, which is of course impossible.