Problem source

Show that $\sin A = \cos B$ if and only if $A = (4n+1)\frac{\pi}{2}\pm B$ for some integer $n$.
Show also that $\big\vert\sin x \pm \cos x \big\vert \le \sqrt{2}$ for all values of $x$ and deduce  that there are no solutions to the equation $\sin\left( \sin x \right) = \cos \left( \cos x \right)$.
Sketch, on the same axes, the graphs of $y= \sin \left( \sin x \right)$ and $y = \cos \left( \cos x \right)$. Sketch, not on the previous axes, the graph of  $y= \sin \left(2 \sin x \right)$.

Solution source

\begin{align*}
&& \sin A &= \cos B \\
\Leftrightarrow && 0 &= \sin A - \cos B \\
&&&=  \sin A - \sin ( \frac{\pi}{2} - B) \\
&&&= 2 \sin \left ( \frac{A + B - \frac{\pi}{2}}{2} \right) \cos \left  (\frac{A - B + \frac\pi2}{2} \right) \\
\Leftrightarrow && n \pi &= \frac{A+B - \frac{\pi}{2}}{2}, n\pi + \frac{\pi}{2} = \frac{A-B+\frac{\pi}{2}}{2} \\
\Leftrightarrow && A \pm B &= 2n\pi + \frac{\pi}{2} \\
&&&= (4n+1) \frac{\pi}{2}
\end{align*}

\begin{align*}
|\sin x \pm \cos x| &= | \sqrt{2} \sin(x \pm \frac{\pi}{4} )| \\
& \leq \sqrt{2}
\end{align*}

Therefore if $\sin(\sin x) = \cos (\cos x)$ we must have that $|\sin x \pm \cos x| = |(4n+1) \frac{\pi}{2}| \geq \frac{\pi}{2} > 1.5 > \sqrt{2}$ contradiction.

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sin(deg(sin(deg(#1))))};
    \def\functiong(#1){cos(deg(cos(deg(#1))))};
    \def\xl{-10};
    \def\xu{10};
    \def\yl{-3};
    \def\yu{3};
    
    % 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
    }
    % 3 - 9/x + 27/x^2 - 147/(2 x^3) + 198/x^4 - 4275/(8 x^5) + 1443/x^6 - 62343/(16 x^7) + 10521/x^8 - 3636267/(128 x^9) + O((1/x)^10)
% (Laurent series)
    % 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[thick, blue, smooth, domain=-10:10, samples=100] 
            plot ({\x}, {\functionf(\x)});
        \draw[thick, green, smooth, domain=-10:10, samples=100] 
            plot ({\x}, {\functiong(\x)});

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

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sin(2*deg(sin(deg(#1))))};
    % \def\functiong(#1){cos(deg(cos(deg(#1))))};
    \def\xl{-10};
    \def\xu{10};
    \def\yl{-3};
    \def\yu{3};
    
    % 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
    }
    % 3 - 9/x + 27/x^2 - 147/(2 x^3) + 198/x^4 - 4275/(8 x^5) + 1443/x^6 - 62343/(16 x^7) + 10521/x^8 - 3636267/(128 x^9) + O((1/x)^10)
% (Laurent series)
    % 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[thick, blue, smooth, domain=-10:10, samples=100] 
            plot ({\x}, {\functionf(\x)});
        % \draw[thick, green, smooth, domain=-10:10, samples=100] 
        %     plot ({\x}, {\functiong(\x)});

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