Let $m$ be an odd and $n$ an even positive integers. I need to estimate the maximum value of $\left|\sin(m\theta)\sin(n\theta)\right|$: $$ c_{m,n}:=\max_{\theta\in[0,\pi]} \left| \sin(m\theta) \sin(n\theta) \right| $$ It is not difficult to show that $c_{1,n}\leq \cos\left(\frac{\pi}{2(n+1)}\right)$. I would like to see if this sort of inequality is true for $c_{m,n}$.
Any answer or reference will be appreciated.
$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\renewcommand{\th}{\theta}$The upper bound $\exp-\frac{d^2}{2(m^2+n^2)}$ on $c_{m,n}$, obtained in the previous answer, can be improved as follows: \begin{equation*} c_{m,n}\le\cos\frac{\pi d}{2(m+n)}, \tag{100}\label{100} \end{equation*} where $d$ is the greatest common divisor of $m$ and $n$.
Indeed, reasoning as in the previous answer (cf. (10), (20), (40), and (50) there) and using the substitutions \begin{equation*} u:=\frac n{m+n}\in(0,1),\quad \th:=\frac{\pi d}{2(m+n)}, \end{equation*} and also using the interchangeability of $m$ and $n$, we reduce \eqref{100} to \begin{equation*} (1+\cos\ep)(1+\cos\de)\le4\cos^2\th \tag{110}\label{110} \end{equation*} given that \begin{equation*} 0\le\ep\le\de\le\pi,\quad u\ep+(1-u)\de\ge2\th. \tag{120}\label{120} \end{equation*}
But \eqref{120} implies \begin{equation*} (1+\cos\ep)(1+\cos\de)\le2(1+\cos\de)=4\cos^2\tfrac\de2 \le4\cos^2\th, \end{equation*} since $0\le\th\le u\frac\ep2+(1-u)\frac\de2\le\tfrac\de2\le\frac\pi2$. So, \eqref{110} follows. $\quad\Box$
$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$Note that \begin{equation*} c_{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):=\tfrac14\,(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=\pi\tfrac{2a+\al}{2m}-\al\tfrac\ep{2m}=\pi\tfrac{2b+\be}{2n}-\be\tfrac\de{2n}
\end{equation*}
for some $\al$ and $\be$ in the set $\{-1,1\}$.
So, by \eqref{30} and the triangle inequality,
\begin{equation*}
h\le\pi|\tfrac{2a+\al}{2m}-\tfrac{2b+\be}{2n}|=|\al\tfrac\ep{2m}-\be\tfrac\de{2n}|
\le\tfrac\ep{2m}+\tfrac\de{2n},
\end{equation*}
so that
\begin{equation*}
\frac\ep{2m}+\frac\de{2n}\ge h. \tag{50}\label{50}
\end{equation*}
Using now the inequality
\begin{equation*}
1-\cos u\le2-2(1-|u|/\pi)^2
\end{equation*}
for $|u|\le\pi$, and recalling \eqref{40} and \eqref{50},
we see that
\begin{equation*}
f(t)\le(1-\ep^2/\pi^2)(1-\de^2/\pi^2)\le\exp-\frac{\ep^2+\de^2}{\pi^2} \\
\le\exp-\frac{4h^2m^2n^2}{\pi^2(m^2+n^2)}=\exp-\frac{d^2}{m^2+n^2}.
\end{equation*}
Thus, by \eqref{10} and \eqref{30},
\begin{equation*}
c_{m,n}\le\exp-\frac{d^2}{2(m^2+n^2)},
\end{equation*}
where $d$ is the greatest common divisor of $m$ and $n$.
