Problem source

By considering the graphs of $y=kx$ and $y=\sin x,$ show that the
equation $kx=\sin x,$ where $k>0,$ may have $0,1,2$ or $3$ roots
in the interval $(4n+1)\frac{\pi}{2} < x < (4n+5)\frac{\pi}{2},$
where $n$ is a positive integer. 

For a certain given value of $n$, the equation has exactly one root
in this interval. Show that $k$ lies in an interval which may be
written $\sin\delta  < k < \dfrac{2}{(4n+1)\pi},$ where $0 < \delta < \frac{1}{2}\pi$
and 
\[
\cos\delta=\left((4n+5)\frac{\pi}{2}-\delta\right)\sin\delta.
\]
Show that, if $n$ is large, then $\delta\approx\dfrac{2}{(4n+5)\pi}$
and obtain a second, improved, approximation.

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sin(deg(#1))};
    \def\xl{-.2};
    \def\xu{20};
    \def\yl{-1.5};
    \def\yu{1.5};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[dashed, blue, smooth, domain=\xl:\xu, samples=100] 
            plot ({\x}, {\functionf(\x)});
        \draw[thick, red, smooth, domain=(5*pi/2):(9*pi/2), samples=100] 
            plot ({\x}, {\functionf(\x)});

        \draw (0,0) -- (\xu, 1);
        \draw (0,0) -- (\xu, 4);
        \draw (0,0) -- (\xu, 2);
        \draw (0,0) -- (\xu, {2*\xu/(9*pi)});
    \end{scope}


    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$t$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}


Clearly we can achieve $0$, $1$, and $2$ intersections by never entering the range, entering too flat, or entering before hitting the second branch. To achieve $3$ we can go at a flat enough slope that we hit somewhere near the top of the second branch, and since the gradient there will be $\approx 0$, and our gradient is positive, we must intersect before that point as well, ie $3$ intersections. Clearly we cannot intersect the second branch $3$ times or the first branch twice, therefore there are at most $3$ intersections.

To intersect the graph only once, we need to:

\begin{itemize}
\item be below $((4n+1)\tfrac{\pi}{2}, 1)$ and 
\item not touch the second gradient
\end{itemize}

The first condition means that $k (4n+1)\tfrac{\pi}{2} < 1 \Rightarrow k < \frac{2}{(4n+1)\pi}$. For the second condition, consider a point on the curve $\sin x$ whose tangent line goes through the origin, ie $\frac{y - \sin t}{x - t} = \cos t \Rightarrow y = (\cos t)x - t \cos t+\sin t$ ie $\sin t = t \cos t$. For this point $t$ to be in the required interval, we need $(4n+5) \tfrac{\pi}{2} -t \in (0, \frac{\pi}{2})$, so let's call this value $\delta$. Then our result is:

The gradient needs to be steeper than $\cos t = \cos \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) = \sin \delta$ and $\cos \delta  =\left ( (4n+5) \tfrac{\pi}{2} - \delta \right)  \sin \delta $.

If $n$ is large, then,

\begin{align*}
&& 1 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right)  \delta \\
\Rightarrow && 1 &\approx (4n+5) \tfrac{\pi}{2} \delta \\
\Rightarrow && \delta &\approx \frac{2}{(4n+5)\pi}
\end{align*}.

To higher order:

\begin{align*}
&& 1-\frac12 \delta^2 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right)  \delta \\
\Rightarrow && 1-\frac12 \delta^2 &\approx (4n+5) \tfrac{\pi}{2} \delta - \delta^2  \\
\Rightarrow && 0 &\approx 1 - (4n + 5)\tfrac{\pi}{2} \delta + \frac12 \delta^2 \\
\Rightarrow && \delta &\approx (4n+5) \tfrac{\pi}{2} - \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2} \\
&&&= \frac{2}{(4n+5) \tfrac{\pi}{2} + \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2}}
\end{align*}.