2x2 reflection matrices

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2x2 reflection matrices

The line in which the unit square is to be reflected has equation $y = \tan(\theta)x$

reflection 2

the position vectors of $A$ and $B$ can be shown to be $$ A =\left(\begin{array}{c}-\cos(2\theta) \\ \sin(2\theta) \end{array}\right) $$ $$ B = \left(\begin{array}{c}\sin(2\theta) \\ \cos(2\theta) \end{array}\right) $$

It follows that the transformation matrix is $$ R_\theta = \left(\begin{array}{cc}-\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & \cos(2\theta)\end{array}\right) $$

With this matrix now constructed, we can see that $\det R_\theta = -1$. $$ \det R_\theta = -\cos(2\theta)\times \cos(2\theta) - \sin(2\theta)\times \sin(2\theta) = -1 $$

2x2 reflection matrices

Given two reflections, $R_\theta$ and $R_\phi$, show that their product, $R_\phi R_\theta$, is a rotation
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$$R_\phi\times R_\theta = \left(\begin{array}{cc}-\cos(2\phi) & \sin(2\phi) \\ \sin(2\phi) & \cos(2\phi)\end{array}\right)\left(\begin{array}{cc}-\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & \cos(2\theta)\end{array}\right)$$

Multiplying out this pair fo matrices gives

$$ R_\phi\times R_\theta = \left(\begin{array}{cc}\cos(2\phi)\cos(2\theta) + \sin(2\phi)\sin(2\theta) & -\cos(2\phi)\sin(2\theta) + \sin(2\phi)\cos(2\theta)\\ \cos(2\phi)\sin(2\theta) - \sin(2\phi)\cos(2\theta)& \cos(2\phi)\cos(2\theta) + \sin(2\phi)\sin(2\theta) \end{array}\right) $$

Applying the compound angle formulae will now reduce each of the terms to give

$$ R_\phi\times R_\theta = \left(\begin{array}{cc} \cos(2\phi - 2\theta) & \sin(2\phi -2\theta) \\ -\sin(2\phi - 2\theta) & \cos(2\phi - 2\theta) \end{array}\right) $$

Note that we need to manipulate some of the terms using $\cos(x-y) \equiv \cos(y-x)$ and $\sin(x-y) \equiv -\sin(y-x)$

$$ R_\phi\times R_\theta = \left(\begin{array}{cc} \cos(2\theta - 2\phi) & -\sin(2\theta -2\phi) \\ \sin(2\theta - 2\phi) & \cos(2\theta - 2\phi) \end{array}\right) $$