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\begin{questionparts}
\item $x_2$ and $y_2$ are defined in terms of $x_1$ and $y_1$ by the equation
$$\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$
$G_1$ is the graph with equation 
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
and $G_2$ is the graph with equation
$$\frac{\left(\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{9} + \frac{\left(-\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{4} = 1$$
Show that, if $(x_1, y_1)$ is a point on $G_1$, then $(x_2, y_2)$ is a point on $G_2$.
Show that $G_2$ is an anti-clockwise rotation of $G_1$ through $45°$ about the origin.
\item \begin{enumerate}
\item The matrix 
$$\begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix}$$
represents a reflection. Find the line of invariant points of this matrix.
\item Sketch, on the same axes, the graphs with equations 
$$y = 2^x \text{ and } 0.8x + 0.6y = 2^{-0.6x+0.8y}$$
\end{enumerate}
\item Sketch, on the same axes, for $0 \leq x \leq 2\pi$, the graphs with equations 
$$y = \sin x \text{ and } y = \sin(x - 2y)$$
You should determine the exact co-ordinates of the points on the graph with equation $y = \sin(x - 2y)$ where the tangent is horizontal and those where it is vertical.
\end{questionparts}

Solution (Optional)

\begin{questionparts}
\item Suppose \begin{align*}
&& \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \\
\Rightarrow && \binom{x_1}{y_1} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \binom{x_2}{y_2}
\end{align*}
Therefore if $\frac{x_1^2}9+\frac{y_1^2}{4} = 1$ we must have

\begin{align*}
\frac{(\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2 }{9} + \frac{(-\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2}{4} = 1
\end{align*}

but this is precisely the statement that $(x_1, y_1)$ is on $G_1$ is equivalent to $(x_2,y_2)$ being on the $G_2$. Since the point $(x_2,y_2)$ is a $45^{\circ}$ rotation of $(x_1,y_1)$ anticlockwise about the origin, this means $G_2$ is a $45^{\circ}$ anticlockwise rotation of $G_1$.

\item \begin{enumerate}
\item
\begin{align*}
&& \begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\
\Rightarrow && \begin{pmatrix} -0.6 x + 0.8y \\ 0.8x + 0.6y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\
\Rightarrow && \begin{pmatrix} -1.6 x + 0.8y \\ 0.8x -0.4y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\
\Rightarrow && y &=2 x
\end{align*}
\item 
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2^(#1)};
    \def\xl{-7};
    \def\xu{7};
    \def\yl{-7};
    \def\yu{7};
    
    % 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[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, black, dashed] (\xl, 2*\xl) -- (\xu, 2*\xu) node [below left] {$y=2x$};
        \draw[thick, red, smooth, domain=\xl:\xu, samples=100] 
            plot ({-0.6*\x+0.8*\functionf(\x)}, {0.8*\x+0.6*\functionf(\x)});
        % \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        % \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
    \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}
\end{enumerate}
\item Consider the transformation $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ which is a shear, leaving the $x$-axis invariant. Then we must have:

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sin(#1)};
    \def\xl{-1};
    \def\xu{7};
    \def\yl{-2};
    \def\yu{2};
    
    % 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[thick, blue, smooth, domain=0:2*pi, samples=100] 
            plot (\x, {\functionf(deg(\x))});
        \draw[thick, red, smooth, domain=0:2*pi, samples=100] 
            plot ({\x+2*\functionf(deg(\x))}, {\functionf(deg(\x))});
        % \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        % \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
    \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}

Since the shear leaves lines of the form $y = k$ invariant, the points where $\frac{\d y}{\d x} = 0$ must also map to points where this is true, ie $(\tfrac{\pi}{2}, 1), (\tfrac{3\pi}{2}, -1)$ map to points $(\tfrac{\pi}{2}+2,1), (\tfrac{3\pi}{2} -2,-1)$ where the tangent is horizontal.

The line $x = c$ map back to lines $\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix} \begin{pmatrix} c \\ t\end{pmatrix} = \begin{pmatrix}c - 2t \\ t \end{pmatrix}$, ie $y = -\frac12 x- \frac{c}{2}$. Therefore we are interested in points on the original curve where the gradient is $-\frac12$, ie $(\frac{2\pi}{3}, \frac{\sqrt{3}}{2}), (\frac{4\pi}{3}, -\frac{\sqrt{3}}{2})$, these map to $(\frac{2\pi}{3}+\sqrt{3},\frac{\sqrt{3}}{2}),  (\frac{4\pi}{3}-\sqrt{3}, -\frac{\sqrt{3}}{2})$

\end{questionparts}

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