Question

If \[ A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}, \] prove that \[ A^3 = \begin{bmatrix} \cos 3\theta & \sin 3\theta \\ -\sin 3\theta & \cos 3\theta \end{bmatrix}. \]

Hint:

To prove powers of rotation matrices, use angle addition formulas to simplify the results.

Answer 1

Step 1: Find \( A^2 \).
First, compute \( A^2 = A \times A \): \[ A^2 = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \] Multiplying the matrices: \[ A^2 = \begin{bmatrix} \cos^2 \theta - \sin^2 \theta & 2 \sin \theta \cos \theta \\ -2 \sin \theta \cos \theta & \cos^2 \theta - \sin^2 \theta \end{bmatrix} \] We can simplify this using the double angle identities: \[ A^2 = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\-\sin 2\theta & \cos 2\theta \end{bmatrix} \]
Step 2: Find \( A^3 \).
Now, compute \( A^3 = A^2 \times A \): \[ A^3 = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \] Multiplying the matrices: \[ A^3 = \begin{bmatrix} \cos 2\theta \cos \theta - \sin 2\theta \sin \theta & \cos 2\theta \sin \theta + \sin 2\theta \cos \theta \\ -(\cos 2\theta \sin \theta - \sin 2\theta \cos \theta) & -\cos 2\theta \cos \theta + \sin 2\theta \sin \theta \end{bmatrix} \] Using the sum of angles identities: \[ A^3 = \begin{bmatrix} \cos (2\theta + \theta) & \sin (2\theta + \theta) \\ -\sin (2\theta + \theta) & \cos (2\theta + \theta) \end{bmatrix} \] \[ A^3 = \begin{bmatrix} \cos 3\theta & \sin 3\theta \\ -\sin 3\theta & \cos 3\theta \end{bmatrix} \] Thus, we have proven that: \[ A^3 = \begin{bmatrix} \cos 3\theta & \sin 3\theta \\ -\sin 3\theta & \cos 3\theta \end{bmatrix} \]