How prove that $\max(|f(1)|,|f(2)|,|f(3)|,|f(4)|)\geq \frac{1}{2}$ if $f(x) = \cos(Ax)+\cos(Bx)$? Let $ A, B$ be real numbers and $ f(x) =\cos(Ax) + \cos(Bx)$. How prove that
$ \max(|f(1)|,|f(2)|,|f(3)|,|f(4)|)\geq \frac{1}{2}$?
 A: By the Briggs' formulas, $\cos(Ax)+\cos(Bx) = 2\cos(Cx)\cos(Dx)$, where $C=\frac{A+B}{2}$ and $D=\frac{A-B}{2}$. By the multiplication formula for the cosine function (aka Chebyshev polynomials of the first type) everything just depends on $\cos(C)=\cos(u)$ and $\cos(D)=\cos(v)$. The set:
$$ S=\left\{x\in[0,2\pi]:|\cos x|\geq \frac{1}{2}\right\} \tag{1}$$
is the union of the intervals:
$$ S=\left[0,\frac{\pi}{3}\right]\cup\left[\frac{2\pi}{3},\frac{4\pi}{3}\right]\cup\left[\frac{5\pi}{3},2\pi\right]$$
and hence it has relative measure $\frac{2}{3}$ with respect to $[0,2\pi]$. We just need to show that for any choice of $P=(u,v)\in\mathbb{T}\times\mathbb{T}$ at least one point among $P,2P,3P,4P$ belongs to $S\times S$.
For any $\{E,F,G\}\subset\{1,2,3,4\}$ we have:
$$\max(|\cos Eu|,|\cos Fu|,|\cos Gu|)\geq\frac{1}{2},$$

(proof without words), so we can choose $\{G,H\}\in\{1,2,3,4\}$ such that both $|\cos Gu|$ and $|\cos Hu|$ are greater or equal to $\frac{1}{2}$. Then for some $I\in\{G,H\}$ we have $|\cos Iv|\geq\frac{1}{2}$, hence we have found a point $x\in\{1,2,3,4\}$ in which the absolute value of $2\cos(Cx)\cos(Dx)$ is greater or equal to $2\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{2}$, as wanted.
This is the (very irregular) graphics of $\max(|f(1)|,|f(2)|,|f(3)|,|f(4)|)$ as a function of $A$ and $B$:

