Prove: $(\forall m, n\in\Bbb N_{>0})(\exists x\in\Bbb R)$ s. t. $2\sin n x \cos m x \ge 1$

Question

Problem 1: Prove that for any $m,n\in\Bbb N_{>0}$, there exists $x \in\Bbb R$ such that $2\sin n x \cos m x \ge 1$.

Four months ago, someone asked the above question. However, when I wanted to post my answer, the question was deleted. I searched by Approach0 without any result. I think that it is a nice question. I don't know why it was deleted. I post it here. I don't remember who posted it before.

Edit 2021/02/20: We may restrict $x$ to be the form $x = r\pi$ where $r$ is a rational number. We give the following problem:

Problem 2: Prove that for any $m, n \in \mathbb{N}_{>0}$, there exists rational $r$ such that $2\sin (n r \pi) \cos (m r\pi) \ge 1$.

Any comments and solutions are welcome and appreciated.

Partial results are as follows.

If $n = m$, let $x = \frac{\pi}{4n}$ and we have $2\sin n x \cos m x = \sin 2n x = 1$.

If $n > m$, let $x = \frac{\pi}{2(2n-1)}$. Since $0 < n x < \pi$ and $0 < m x \le (n-1)x < \pi$, we have \begin{align} 2\sin n x \cos m x &\ge 2\sin n x \cos (n-1)x \\ &= \sin (2n-1)x + \sin x \\ &= 1 + \sin \frac{\pi}{2(2n-1)}\\ & \ge 1. \end{align}

Answer 1

hint

Using the transformation formula, it will be equivalent to prove that

$$\sin((n+m)x)+\sin((n-m)x)\ge \color{red}{1}$$

Assume that $n\ge m>0$, we can take $$x=\frac{\pi}{2(n+m)}$$

then $$\sin((n+m)x)=\color{red}{1}$$ and $$0\le (n-m)x<\frac{\pi}{2}$$

so $$\sin((n-m)x)\ge\color{red}{0}$$

Answer 2

This answer shows that, for any given two positive integers $m,n$ with $m\gt n$, there exists a rational number $r$ such that $$2\sin(nr\pi)\cos(mr\pi)\geq 1$$

Proof :

We see that $$2\sin(nx)\cos(mx)\ge 1$$ is equivalent to $$\sin((m+n)x)-\sin((m-n)x)\ge 1\tag1$$

To get $x$ satisfying $(1)$, let us consider $x$ such that $$\sin((m+n)x)=1\qquad\text{and}\qquad \sin((m-n)x)\le 0\tag2$$

We have $$\sin((m+n)x)=1\iff (m+n)x=\frac{\pi}{2}+2s\pi\iff x=\frac{4s+1}{2(m+n)}\pi$$ where $s$ is an integer. So, we get $$\begin{align}(2)&\iff \pi+2t\pi\le (m-n)\frac{4s+1}{2(m+n)}\pi\le 2\pi+2t\pi \\\\&\iff 4t+2\le (4s+1)\frac{m-n}{m+n}\le 4t+4\end{align}$$ where $t$ is an integer.

For any given $m,n$ with $m\gt n$, we want to find a pair of integers $(s,t)$ such that $$4t+2\le f(s)\le 4t+4$$ where $$f(s):=(4s+1)\frac{m-n}{m+n}$$

We know that $s=0$ doesn't work since $0\lt f(0)=\dfrac{m-n}{m+n}\lt 1$.

If either $s=1$ or $s=2$ works, then we are happy.

In the following, let us consider the case where $s=1,2$ don't work.

• Suppose that $8\lt f(1)$. Then, we have $3m+13n\lt 0$ which is impossible. So, we get $f(1)\le 8$.

• Suppose that $12\lt f(2)$. Then, we have $m+7n\lt 0$ which is impossible. So, we get $f(2)\le 12$.

• Suppose that $0\lt f(1)\lt 2$ and $4\lt f(2)$. Then, we have $\frac 37\lt \frac nm\lt \frac{5}{13}$ which is impossible. So, if $0\lt f(1)\lt 2$, then $0\lt f(2)\lt 2$.

• Suppose that $4\lt f(1)\lt f(2)\lt 6$. Then, we have $\frac 15\lt \frac nm\lt \frac 19$ which is impossible. So, if $4\lt f(1)\lt 6$, then $8\lt f(2)\lt 10$.

So, we have two cases to consider :

Case 1 : $0\lt f(1)\lt f(2)\lt 2$

We have $f(2)\lt 2\implies \dfrac{7}{11}\lt \dfrac nm$ which implies $$0\lt f(s+1)-f(s)=\frac{4(m-n)}{m+n}=\frac{4-4\frac nm}{1+\frac nm}\lt \frac{4-4\cdot \frac{7}{11}}{1+\frac{7}{11}}=\frac 89\lt 1$$ So, it is sufficient to find the smallest integer $s$ such that $f(s)\ge 2$, i.e. $\dfrac{m+3n}{4(m-n)}\le s$. So, we see that $s=\left\lceil \dfrac{m+3n}{4(m-n)}\right\rceil$ works where we can take $t=0$ since $2\le f(s)\le 4$.

Case 2 : $4\lt f(1)\lt 6$ and $8\lt f(2)\lt 10$

We have $\dfrac{m}{n}\gt 17$ which implies $$3=4-1\lt f(1)-f(0)= f(s+1)-f(s)=\frac{4+4\frac mn}{\frac mn-1}\lt \frac{4+4\times 17}{17-1}=\frac 92$$ So, it is sufficient to find the smallest $s$ such that $f(s)\le 4s$, i.e. $\dfrac{m-n}{8n}\le s$. So, we see that $s=\left\lceil\dfrac{m-n}{8n}\right\rceil$ works where we can take $t=s-1$ since $4(s-1)+2\le f(s)\le 4(s-1)+4$.

It follows from these that we can take $$x=\begin{cases}\dfrac{5}{2(m+n)}\pi&\text{if $s=1$ works} \\\\\dfrac{9}{2(m+n)}\pi&\text{if $s=1$ doesn't work, and $s=2$ works} \\\\\dfrac{4\left\lceil \dfrac{m+3n}{4(m-n)}\right\rceil+1}{2(m+n)}\pi&\text{if $s=1,2$ don't work, and $f(2)\lt 2$} \\\\\dfrac{4\left\lceil\dfrac{m-n}{8n}\right\rceil+1}{2(m+n)}\pi&\text{otherwise}\end{cases}$$ which can be simplified as $$x=\begin{cases}\dfrac{5}{2(m+n)}\pi&\text{if $\ \dfrac nm\in \bigg[\dfrac 19,\dfrac 37\bigg]$} \\\\\dfrac{9}{2(m+n)}\pi&\text{if $\ \dfrac nm\in \bigg[\dfrac{1}{17},\dfrac 19\bigg)\cup\bigg(\dfrac 37,\dfrac{7}{11}\bigg]$} \\\\\dfrac{4\left\lceil \dfrac{m+3n}{4(m-n)}\right\rceil+1}{2(m+n)}\pi&\text{if $\ \dfrac nm\in \bigg(\dfrac{7}{11},1\bigg)$} \\\\\dfrac{4\left\lceil\dfrac{m-n}{8n}\right\rceil+1}{2(m+n)}\pi&\text{if $\ \dfrac nm\in\bigg(0,\dfrac{1}{17}\bigg)$}\end{cases}$$

Answer 3

Remarks: Here is my proof for Problem 2. I wrote it for Problem 1 originally, though Problem 1 might be easier. By the way, @mathlove gave a nice proof.

Problem 2: Prove that for any $m, n \in \mathbb{N}_{>0}$, there exists rational $r$ such that $2\sin (n r \pi) \cos (m r\pi) \ge 1$.

Proof: Let $x = r\pi$. We split into five cases:

1. $n > m$: Let $x = \frac{\pi}{2(2n-1)}$. Since $0 < n x < \pi$ and $0 < m x \le (n-1)x < \pi$, we have \begin{align} 2\sin n x \cos m x &\ge 2\sin n x \cos (n-1)x \\ &= \sin (2n-1)x + \sin x \\ &= 1 + \sin \frac{\pi}{2(2n-1)}\\ & \ge 1. \end{align}

2. $n = m$: Let $x = \frac{\pi}{4n}$. We have $2\sin n x \cos m x = \sin 2n x = 1$.

3. $\frac{5m}{6} < n < m$: Let $x = -\frac{3\pi}{4m}$. Since $\frac{5\pi}{8} < \frac{3n\pi}{4m} < \frac{3\pi}{4}$, we have \begin{align} 2\sin n x \cos mx &= \sqrt{2}\, \sin \frac{3n\pi}{4m}\\ &\ge \sqrt{2}\, \sin \frac{3\pi}{4}\\ & = 1. \end{align}

4. $\frac{m}{6} \le n \le \frac{5m}{6}$: Let $x = -\frac{\pi}{m}$. Since $\frac{\pi}{6} \le \frac{n\pi}{m} \le \frac{5\pi}{6}$, we have $$2\sin n x \cos m x = 2\sin \frac{n\pi}{m} \ge 2\sin \frac{\pi}{6} = 1.$$

5. $n < \frac{m}{6}$: Let $$y = \frac{\pi}{m}\Big\lfloor \frac{1}{2}\frac{m}{n} + \frac{1}{2}\Big\rfloor$$ where $\lfloor \cdot \rfloor$ is the floor function.

Clearly, $|\cos my| = 1$. By using $z-1 < \lfloor z \rfloor \le z$ and $\frac{n}{m} < \frac{1}{6}$, we have $$ny \ge \frac{n\pi}{m}\Big(\frac{1}{2}\frac{m}{n} + \frac{1}{2} - 1\Big) = \frac{\pi}{2} - \frac{n\pi}{2m} \ge \frac{5\pi}{12}$$ and $$n y \le \frac{n\pi}{m}\Big(\frac{1}{2}\frac{m}{n} + \frac{1}{2}\Big) = \frac{\pi}{2} + \frac{n\pi}{2m} < \frac{7\pi}{12}.$$ Thus, we have $\sin ny \ge \sin \frac{5\pi}{12}$ and thus $$|2\sin ny \cos my| \ge 2 \sin \frac{5\pi}{12} > 1.$$ If $2\sin ny \cos my > 0$, let $x = y$ and we have $2\sin n x \cos mx > 1$.

If $2\sin ny \cos my < 0$, let $x = -y$ and we have $2\sin n x \cos mx > 1$.

We are done.