QCAA Specialist Mathematics Alternative Sequence Matrices and transformations

Method 1: Assumes the rotation through A as the positive direction.
First rotation matrix:
$\mathbf{R}_A = \begin{bmatrix} \cos(A) & -\sin(A) \\ \sin(A) & \cos(A) \end{bmatrix}$

Second rotation matrix:
$\mathbf{R}_{-B} = \begin{bmatrix} \cos(-B) & -\sin(-B) \\ \sin(-B) & \cos(-B) \end{bmatrix}$
$= \begin{bmatrix} \cos(B) & \sin(B) \\ -\sin(B) & \cos(B) \end{bmatrix}$

Matrix representing the composition of the successive rotations:
$\mathbf{R}_{-B} \mathbf{R}_A = \begin{bmatrix} \cos(B) & \sin(B) \\ -\sin(B) & \cos(B) \end{bmatrix} \begin{bmatrix} \cos(A) & -\sin(A) \\ \sin(A) & \cos(A) \end{bmatrix}$
$= \begin{bmatrix} \cos(B)\cos(A) + \sin(B)\sin(A) & -\cos(B)\sin(A) + \sin(B)\cos(A) \\ -\sin(B)\cos(A) + \cos(B)\sin(A) & \sin(B)\sin(A) + \cos(B)\cos(A) \end{bmatrix}$

Matrix representing transformation of the combined rotation of $A - B$
$\mathbf{R}_{A-B} = \begin{bmatrix} \cos(A - B) & -\sin(A - B) \\ \sin(A - B) & \cos(A - B) \end{bmatrix}$

$\mathbf{R}_{-B} \mathbf{R}_A$ and $\mathbf{R}_{A-B}$ represent the same transformation.

Equating parts:
$\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$
$\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)$

Rotation matrix
$T_1=\begin{bmatrix}\cos\left(-\frac{\pi}{2}\right) & -\sin\left(-\frac{\pi}{2}\right) \\ \sin\left(-\frac{\pi}{2}\right) & \cos\left(-\frac{\pi}{2}\right)\end{bmatrix}=\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$

Reflection matrix
$T_2=\begin{bmatrix}\cos\left(2\times\frac{\pi}{8}\right) & \sin\left(2\times\frac{\pi}{8}\right) \\ \sin\left(2\times\frac{\pi}{8}\right) & -\cos\left(2\times\frac{\pi}{8}\right)\end{bmatrix}=\begin{bmatrix}\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\end{bmatrix}$

$\begin{bmatrix}x'\\y'\end{bmatrix}=T_2T_1\begin{bmatrix}x\\y\end{bmatrix}\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{bmatrix}^{-1}\begin{bmatrix}-2\\4\end{bmatrix}$
$\begin{bmatrix}x\\y\end{bmatrix}=\frac{\sqrt{2}}{2}\begin{bmatrix}6\\2\end{bmatrix}$
The coordinates of $P$ are $(3\sqrt{2},\sqrt{2})$

Gradient of $PP'$ $=\frac{\sqrt{2}-4}{3\sqrt{2}+2}$
$=\frac{(\sqrt{2}-4)(3\sqrt{2}-2)}{(3\sqrt{2}+2)(3\sqrt{2}-2)}$
$=\frac{14(1-\sqrt{2})}{14}$
$=1-\sqrt{2}$