Let $A_1 , A_2 , w_1, w_2 $ are positives numbers If for all real t , $s_1(t)=A_1\cos(w_1t+\phi_1)$ , $s_2(t)=A_2\cos(w_2t+\phi_2)$ and $s=s_1 +s_2$
We know if $w_1 = w_2$, we can write $s$ as $s(t)=A\cos(wt+\phi), \forall t\in \mathbb R$.
If $w_1\neq w_2$, I'm trying to prove that, we can't write $s$ as $s(t)=A\cos(wt+\phi), \forall t\in \mathbb R$.
My attempt was deleted (any comment)
A Rigorous Mathematical Proof
If $\forall t > 0, A\cos(wt + \phi) = A_1\cos(w_1t + \phi_1) + A_2\cos(w_2t + \phi_2)$, then for $\forall x > 0$, \begin{equation} A \int_0^x \cos^2(wt + \phi)\, dt = A_1\int_0^x \cos(w_1t + \phi_1)\cos(wt + \phi)\, dt + A_2\int_0^x \cos(w_2t + \phi_2)\cos(wt + \phi)\, dt. \end{equation}
But, \begin{align*} A \int_0^x \cos^2(wt + \phi)\, dt &= A \int_0^x \frac{1+\cos(2wt + 2\phi)}{2}\, dt \\ &= \frac{A}{2} x + g_1(x), \end{align*} with a bounded function $g_1$ on $\mathbb{R}^*_+$.
If $w \neq w_1$ and $w \neq w_2$, then \begin{equation*} A_1\int_0^x \cos(w_1t + \phi_1)\cos(wt + \phi)\, dt + A_2\int_0^x \cos(w_2t + \phi_2)\cos(wt + \phi)\, dt = g_2(x), \end{equation*} with a bounded function $g_2$ on $\mathbb{R}^*_+$. We deduce that $\forall x > 0, \frac{A}{2} x + g_1(x) = g_2(x)$. This is possible only if $A = 0$, leading to a contradiction. Therefore, $w = w_1$ or $w = w_2$.
Now, if, for example, $w = w_1$, we have $$\forall t > 0, A\cos(wt + \phi) - A_1\cos(wt + \phi_1) = A_2\cos(w_2t + \phi_2)$$ But the fundamental period of $A\cos(wt + \phi) - A_1\cos(wt + \phi_1)$ is known as $\frac{2\pi}{w}$, and the fundamental period of $A_2\cos(w_2t + \phi_2)$ is $\frac{2\pi}{w_2}$. Therefore, we conclude that $w = w_2$.
The proof generalizes for the sum of n sinusoidal signals.