# Plane Reflection is Involution

## Theorem

Let $M$ be a straight line in the plane passing through the origin.

Let $s_M$ be the reflection of $\R^2$ in $M$.

Then $s_M$ is an involution in the sense that:

$s_M \circ s_M = I_{\R^2}$

where $I_{\R^2}$ is the identity mapping on $\R_2$.

That is:

$s_M = {s_M}^{-1}$

## Proof

Let the angle between $M$ and the $x$-axis be $\alpha$.

Let $P = \tuple {x, y}$ be an arbitrary point in the plane.

Then from Equations defining Plane Reflection:

$\map {s_M} P = \tuple {x \cos 2 \alpha + y \sin 2 \alpha, x \sin 2 \alpha - y \cos 2 \alpha}$

Thus:

 $\ds \map {s_M \circ s_M} P$ $=$ $\ds \map {s_M} {\map {s_M} P}$ Definition of Composition of Mappings $\ds$ $=$ $\ds \map {s_M} {\tuple {x \cos 2 \alpha + y \sin 2 \alpha, x \sin 2 \alpha - y \cos 2 \alpha} }$ $\ds$ $=$ $\ds {\tuple {\paren {x \cos 2 \alpha + y \sin 2 \alpha} \cos 2 \alpha + \paren {x \sin 2 \alpha - y \cos 2 \alpha} \sin 2 \alpha, \paren {x \cos 2 \alpha + y \sin 2 \alpha} \sin 2 \alpha - \paren {x \sin 2 \alpha - y \cos 2 \alpha} \cos 2 \alpha} }$ $\ds$ $=$ $\ds \tuple {x \cos^2 2 \alpha + y \sin 2 \alpha \cos 2 \alpha + x \sin^2 2 \alpha - y \sin 2 \alpha \cos 2 \alpha, x \sin 2 \alpha \cos 2 \alpha + y \sin^2 2 \alpha - x \sin 2 \alpha \cos 2 \alpha + y \cos^2 2 \alpha}$ $\ds$ $=$ $\ds \tuple {x \paren {\cos^2 2 + \sin^2 2 \alpha}, y \paren {\sin^2 2 \alpha + \cos^2 2 \alpha} }$ simplifying $\ds$ $=$ $\ds \tuple {x, y}$ Sum of Squares of Sine and Cosine

Hence the result.

$\blacksquare$