Year 11 Maths - Specialist Transformations of the plane

Rotations and general reflections

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Questions

Question 1

61443

Find the matrix for each of the following rotations about the origin

i) $60^\circ$ anticlockwise

ii) $225^\circ$ clockwise

Answer

Part a:

$\begin{bmatrix}\tfrac{1}{2}&\tfrac{\sqrt{3}}{2}\\-\tfrac{\sqrt{3}}{2}&\tfrac{1}{2}\end{bmatrix}$

$\begin{bmatrix}\tfrac{\sqrt{3}}{2}&-\tfrac{1}{2}\\\tfrac{1}{2}&\tfrac{\sqrt{3}}{2}\end{bmatrix}$

$\begin{bmatrix}\tfrac{1}{2}&\tfrac{\sqrt{3}}{2}\\\tfrac{\sqrt{3}}{2}&\tfrac{1}{2}\end{bmatrix}$

$\begin{bmatrix}\tfrac{1}{2}&-\tfrac{\sqrt{3}}{2}\\\tfrac{\sqrt{3}}{2}&\tfrac{1}{2}\end{bmatrix}$

Correct

Part b:

$\begin{bmatrix}\tfrac{1}{\sqrt{2}}&\tfrac{1}{\sqrt{2}}\\-\tfrac{1}{\sqrt{2}}&\tfrac{1}{\sqrt{2}}\end{bmatrix}$

$\begin{bmatrix}-\tfrac{1}{\sqrt{2}}&-\tfrac{1}{\sqrt{2}}\\\tfrac{1}{\sqrt{2}}&-\tfrac{1}{\sqrt{2}}\end{bmatrix}$

Correct

$\begin{bmatrix}\tfrac{1}{\sqrt{2}}&-\tfrac{1}{\sqrt{2}}\\\tfrac{1}{\sqrt{2}}&\tfrac{1}{\sqrt{2}}\end{bmatrix}$

$\begin{bmatrix}-\tfrac{1}{\sqrt{2}}&\tfrac{1}{\sqrt{2}}\\-\tfrac{1}{\sqrt{2}}&-\tfrac{1}{\sqrt{2}}\end{bmatrix}$

Worked Solution

$$\begin{align}&\text{(i)}\quad
M=\left[\begin{array}{cc}
\cos 60^{\circ}-\sin 60^{\circ} \\
\sin 60^{\circ} & \cos 60^{\circ}
\end{array}\right]=\left[\begin{array}{cc}
\frac{1}{2} & -\frac{\sqrt{3}}{2} \\
\frac{\sqrt{3}}{2} & \frac{1}{2}
\end{array}\right]\\
&\begin{aligned}\text{(ii)}\quad M &=\left[\begin{array}{c}\cos \left(-225^{\circ}\right) & -\sin \left(-225^{4}\right) \\ \sin \left(-225^{\circ}\right) & \cos \left(-225^{\circ}\right)\end{array}\right] \\ &=\left[\begin{array}{c}\cos \left(225^\circ+\sin\left(225^{\circ}\right)\right. \\ -\sin \left(225^{\circ}\right)+\cos \left(125^{\circ}\right)\end{array}\right] \\ &=\left[\begin{array}{l}-\cos 45^{\circ}-\sin 45^{\circ} \\ +\sin 45^{\circ}-\cos 45^{\circ}\end{array}\right]\\
&=\left[\begin{array}{c}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\end{array}\right] \end{aligned}
\end{align}$$

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