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'Reflect Again' printed from https://nrich.maths.org/
Andrei Lazanu proved the double angle formulae illustrated in the
diagram:
The diagram starts from a right angled triangle, of sides $2t$ and
2 and where consequently $\tan\theta = t$. In this triangle, a line
making an angle $\theta$ with the hypotenuse is drawn. This way, an
isosceles triangle is formed, and $2\theta$ is the angle exterior
to this isosceles triangle. Let the sides DA and DB of this
isosceles triangle be $x$\ units. Then the length of DC must be
$2-x$ units. Using Pythagoras' Theorem for triangle ADC we find
$x$. $$x^2=(2t)^2+(2-x)^2.$$ Hence $x=1+t^2$ and so the length of
side DC is $2-(1+t^2)=1-t^2$.
The formulae for the sine, cosine and tangent of $2\theta$ in terms
of $t$, where $t=\tan \theta$, follow directly from the ratios of
the sides of the right angled triangle ADC and we get $$\tan2\theta
= {2t\over {1-t^2}},\quad \sin2\theta = {2t\over {1+t^2}},\quad
\cos2\theta = {{1-t^2}\over {1+t^2}}$$
Finally, to find the combined transformation you have to
multiply the transformation matricies to get (using the double
angle formulae to simplify):
$$T= \left( \begin{array}{cc} cos 2(\phi-\theta) &
-sin2(\phi-\theta)\\ sin2(\phi-\theta & \cos2(\phi-\theta)
\end{array} \right) $$
Which is a rotation. If the angles are equal the matrix would
be:
$$T= \left( \begin{array}{cc} 1 & 0\\ 0 & 1
\end{array} \right) $$
So the two reflections combined would leave all points unchanged.