Supremum of an expression involving $\sin$ Fix $n,m$ two different positive integers. If $n,m$ are both sufficiently big, does
$$
\sup_{x \in \mathbb{R}} |\sin(nx)-\sin(mx)| = 2 \,\ ?
$$
I've tried to come up with some sequence $(x_k)_{k \geq 1}$ that would show this, but I did not succeed.
 A: Using the difference of sines formula, we get
$$
\sin(nx)-\sin(mx)=2\sin\left(\frac{n-m}2x\right)\cos\left(\frac{n+m}2x\right)\tag{1}
$$
For $m^2\ne n^2$, we would like to arrange
$$
\frac{n-m}{2\pi}x\equiv\frac12\pmod{1}\tag{2}
$$
and
$$
\frac{n+m}{2\pi}x\equiv0\pmod{1}\tag{3}
$$
for then, $|\sin(nx)-\sin(mx)|=2$.
To satisfy $(2)$ and $(3)$, we need some $x\in\mathbb{R}$ and $j,k\in\mathbb{Z}$ so that
$$
(n-m)x=(2j+1)\pi\tag{4}
$$
and
$$
(n+m)x=2k\pi\tag{5}
$$
Thus,
$$
\frac{n+m}{n-m}=\frac{2k}{2j+1}\tag{6}
$$
Therefore, $(2)$ and $(3)$ have a solution if $n+m$ has more factors of $2$ than $n-m$.

Suppose $n+m$ has no more factors of $2$ than $n-m$. Then
$$
\frac{n+m}{n-m}=\frac{2j+1}{k}\tag{7}
$$
Furthermore, suppose
$$
\left|\,\frac{n-m}{2\pi}x-\frac{2p+1}2\,\right|\le\epsilon\tag{8}
$$
Then
$$
\begin{align}
\left|\,\frac{n+m}{2\pi}x-\frac{2j+1}{k}\frac{2p+1}2\,\right|
&=\left|\,\frac{n+m}{2\pi}x-\frac{n+m}{n-m}\frac{2p+1}2\,\right|\\
&\le\frac{n+m}{n-m}\epsilon\tag{9}
\end{align}
$$
and therefore, the triangle inequality and $(9)$ imply
$$
\begin{align}
\left|\,\frac{n+m}{2\pi}x-q\,\right|
&\ge\left|\,q-\frac{2j+1}{k}\frac{2p+1}2\,\right|-\left|\,\frac{n+m}{2\pi}x-\frac{2j+1}{k}\frac{2p+1}2\,\right|\\
&\ge\frac1{2k}-\frac{n+m}{n-m}\epsilon\tag{10}
\end{align}
$$
Therefore,
$$
\left|\,\frac{n-m}{2\pi}x-\frac{2p+1}2\,\right|\le\frac{n-m}{4kn}
\implies\left|\,\frac{n+m}{2\pi}x-q\,\right|\ge\frac{n-m}{4kn}\tag{11}
$$
Applying $(11)$ to $(1)$, we get
$$
|\sin(nx)-\sin(mx)|\le2\cos\left(\frac{n-m}{4kn}\pi\right)\tag{12}
$$

Summary
If $n+m$ has more factors of $2$ than $n-m$, there is an $x$ so that $|\sin(nx)-\sin(mx)|=2$. Otherwise, $|\sin(nx)-\sin(mx)|$ is bounded away from $2$.

Example 1:
If $n=6$ and $m=2$, then $8$ has one more factor of $2$ than $4$. Plugging into $(4)$, $(5)$, and $(6)$ yields $j=0$, $k=1$, and $x=\frac\pi4$:
$$
\left|\,\sin\left(6\cdot\frac\pi4\right)-\sin\left(2\cdot\frac\pi4\right)\,\right|=2
$$

Example 2:
If $n=2$ and $m=1$, then $3$ has no more factors of $2$ than $1$. Plugging into $(7)$ yields $j=1$ and $k=1$. Then $(12)$ says that
$$
\begin{align}
|\sin(2x)-\sin(x)|
&\le2\cos\left(\frac\pi8\right)\\
&\doteq1.84775906502257
\end{align}
$$
Mathematica says that the maximum of $|\sin(2x)-\sin(x)|$ is approximately $1.760172593046$.
A: This is not true. You can see that here for example. What you want to do is write this in a different form:
$$\sin(nx)-\sin(mx) = -2\sin\left(\frac{(m-n)x}{2}\right)\cos\left(\frac{(m+n)x}{2}\right).$$
It's pretty easy to see that it cannot be $2$ from this expression.
