An upper bound on certain finite trigonometric series given a lower bound

Solution 1:

Today,I have solve this problem.

we only note this problem have follow form $$f\left(x-\dfrac{2\pi}{3}\right)+f(x)+f\left(x+\dfrac{2\pi}{3}\right)=3$$

because use $$\sin{(x+y)}+\sin{(x-y)}=2\sin{x}\cos{x}$$ $$\cos{(x+y)}+\cos{(x-y)}=2\cos{x}\cos{y}$$ $$\Longrightarrow \sin{(x-\dfrac{2\pi}{3})}+\sin{(x+\dfrac{2\pi}{3})}=-sin{x}$$ $$\cdots\cdots$$ $$\cos{(2x+\dfrac{4\pi}{3})}+\cos{(2x-\dfrac{4\pi}{3})}=-\cos{(2x)}$$ so $$\sin{(x-\dfrac{2\pi}{3})}+\sin{x}+\sin{(x+\dfrac{2\pi}{3})}=0$$ $$\cos{(x-\dfrac{2\pi}{3})}+\cos{x}+\cos{(x+\dfrac{2\pi}{3})}=0$$ $$\sin{(2x-\dfrac{4\pi}{3})}+\sin{2x}+\sin{(2x+\dfrac{4\pi}{3})}=0$$

$$\cos{(2x-\dfrac{4\pi}{3})}+\cos{2x}+\cos{(2x+\dfrac{4\pi}{3})}=0$$

Solution 2:

As obareey mentioned in the comments to the question, a (finite) sum of sine waves of the same frequency is again a sine wave of that frequency. In particular, we have $$\begin{align} a \sin x + b \cos x &= \sqrt{a^2 + b^2} \cos(x + \phi_1) \\ c \sin 2x + d \cos 2x &= \sqrt{c^2 + d^2} \cos(2x + \phi_2) \end{align}$$ for some $\phi_1, \phi_2 \in \mathbb R$. Denote $\alpha := \sqrt{a^2 + b^2}$, $\beta := \sqrt{c^2 + d^2}$, and $$ g(x) := \alpha \cos(x + \phi_1) + \beta \cos(2x + \phi_2) = f(x) - 1. $$ It suffices to show that if there exists an $x^* \in \mathbb R$ such that $g(x^*) \geq 2$, then there exists an $x' \in \mathbb R$ such that $g(x') \leq -1$.

To prove, first note that since $\alpha, \beta \geq 0$ and $\cos \leq 1$, we have $\alpha + \beta \geq g(x^*) \geq 2$. Next, consider the sets $$\textstyle S := \{ x \in \mathbb{R} : \cos(x + \phi_1) \leq -\frac{1}{2}\} = \bigcup_{k \in \mathbb{Z}} \left[2\pi k - \phi_1 + \frac{2\pi}{3}, 2\pi k - \phi_1 + \frac{4\pi}{3}\right] \\\textstyle T := \{ x \in \mathbb{R} : \cos(2x + \phi_2) \leq -\frac{1}{2}\} = \bigcup_{k \in \mathbb{Z}} \left[\pi k - \frac{\phi_2}{2} + \frac{\pi}{3}, \pi k - \frac{\phi_2}{2} + \frac{2\pi}{3}\right]. $$ Observe that $S$ is composed of closed intervals of length $2\pi/3$, while $$\textstyle \mathbb R \setminus T = \bigcup_{k \in \mathbb{Z}} \left(\pi k - \frac{\phi_2}{2} - \frac{\pi}{3}, \pi k - \frac{\phi_2}{2} + \frac{\pi}{3}\right) $$ is composed of open intervals of length $2\pi/3$. Thus $S \not\subseteq \mathbb{R} \setminus T$, meaning that $S \cap T \neq \varnothing$. If we now choose an $x' \in S \cap T$, then $g(x') \leq -\frac{1}{2}(\alpha + \beta) \leq -1$, as desired.