$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$Note that 
\begin{equation*}
	4c_{m,n}^2=\max_{t\in[0,\pi]}f(t), \tag{10}\label{10}
\end{equation*}
where 
\begin{equation*}
	f(t):=f_{m,n}(t):=(1-\cos2mt)(1-\cos2nt). \tag{20}\label{20}
\end{equation*}
Next, if $\frac{2i+1}{m}=\frac{2j+1}{n}$ for some integers $i$ and $j$, then the even number $(2i+1)n$ equals the odd number $(2j+1)m$, a contradiction. Also, the smallest common denominator for the fractions $\frac{2i+1}{m}$ and $\frac{2j+1}{n}$ is $mn/d$, where $d$ is the greatest common divisor of $m$ and $n$. So, for any integers $i$ and $j$,
\begin{equation*}
	\Big|\frac{2i+1}{2m}\,\pi-\frac{2j+1}{2n}\,\pi\Big|\ge h:=\frac\pi{2mn/d}\in(0,\pi/2]. \tag{30}\label{30}
\end{equation*}

Take any $t\in[0,\pi]$. Then 
\begin{equation*}
	t\in I\cap J,
\end{equation*}
where 
\begin{equation*}
	I:=\pi[\tfrac{2a-1}{2m},\tfrac{2a+1}{2m}],\quad J:=\pi[\tfrac{2b-1}{2n},\tfrac{2b+1}{2n}]
\end{equation*}
for some $a\in\{0,\dots,m\}$ and $b\in\{0,\dots,n\}$, so that 
$|2mt-2\pi a|\le\pi$ and $|2nt-2\pi b|\le\pi$.  
Letting 
\begin{equation*}
	\ep:=\pi-|2mt-2\pi a|\in[0,\pi],\quad\de:=\pi-|2nt-2\pi b|\in[0,\pi], \tag{40}\label{40} 
\end{equation*}
we get 
\begin{equation*}
	t\in\{\tfrac{2a-1}{2m}+\ep,\tfrac{2a+1}{2m}-\ep\}\cap
	\{\tfrac{2b-1}{2n}+\de,\tfrac{2b+1}{2n}-\de\}. 
\end{equation*}
So, by \eqref{30} and the triangle inequality,
\begin{equation}
	\ep+\de\ge h. \tag{50}\label{50} 
\end{equation}
Using now the inequality 
\begin{equation*}
	1-\cos u\le2-c(\pi-|u|)^2
\end{equation*}
for 
\begin{equation*}
	c:=2/\pi^2
\end{equation*}
and $|u|\le\pi$, and recalling \eqref{40} and \eqref{50} and the $\in$ relation in \eqref{30}, 
we see that 
\begin{equation*}
	f(t)=(2-c\ep^2)(2-c\de^2)\le4-2(h/\pi)^2-(h/\pi)^4/4\le4-(31/64)(h/\pi)^2. 
\end{equation*}
Thus, by \eqref{10} and \eqref{30},
\begin{equation}
	c_{m,n}\le\sqrt{1-\frac{31}{256(mn/d)^2}}, 
\end{equation}
where $d$ is the greatest common divisor of $m$ and $n$.