Let `A = [{:("cos"alpha, -"sin"alpha), ("sin"alpha, "cos"alpha"):}], (alpha in R)` such that `A^(32) = [{:(0, -1), (1, 0):}]`. Then, a value of `alpha` is

Options

A. `(pi)/(32)`
B. 0
C. `(pi)/(64)`
D. `(pi)/(16)`

Explanation

To solve the problem, we need to find the value of \(\alpha\) such that the matrix \(A^{32}\) equals \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\).

Given the matrix:
\[ A = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]

### Step 1: Calculate \(A^2\)
To find \(A^{32}\), we first calculate \(A^2\):
\[ A^2 = A \cdot A = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \cdot \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]

Using matrix multiplication:
\[ A^2 = \begin{pmatrix} \cos^2 \alpha - \sin^2 \alpha & -2\sin \alpha \cos \alpha \\ 2\sin \alpha \cos \alpha & \cos^2 \alpha - \sin^2 \alpha \end{pmatrix} \]

### Step 2: Use Trigonometric Identities
We can apply the double angle formulas:
\[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \]
\[ \sin 2\alpha = 2\sin \alpha \cos \alpha \]

Thus, we can express \(A^2\) as:
\[ A^2 = \begin{pmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{pmatrix} \]

### Step 3: Generalize to \(A^n\)
Continuing this pattern, we find that:
\[ A^n = \begin{pmatrix} \cos n\alpha & -\sin n\alpha \\ \sin n\alpha & \cos n\alpha \end{pmatrix} \]

### Step 4: Calculate \(A^{32}\)
Now we can express \(A^{32}\):
\[ A^{32} = \begin{pmatrix} \cos 32\alpha & -\sin 32\alpha \\ \sin 32\alpha & \cos 32\alpha \end{pmatrix} \]

### Step 5: Set Equal to Given Matrix
We know:
\[ A^{32} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \]

This gives us two equations:
1. \(\cos 32\alpha = 0\)
2. \(\sin 32\alpha = 1\)

### Step 6: Solve the Equations
From \(\cos 32\alpha = 0\), we have:
\[ 32\alpha = \frac{\pi}{2} + k\pi \quad (k \in \mathbb{Z}) \]

From \(\sin 32\alpha = 1\), we have:
\[ 32\alpha = \frac{\pi}{2} + 2m\pi \quad (m \in \mathbb{Z}) \]

### Step 7: Equate and Solve for \(\alpha\)
Setting the two expressions for \(32\alpha\) equal:
\[ \frac{\pi}{2} + k\pi = \frac{\pi}{2} + 2m\pi \]

This simplifies to:
\[ k\pi = 2m\pi \implies k = 2m \]

### Step 8: Find \(\alpha\)
Now, we can solve for \(\alpha\):
\[ 32\alpha = \frac{\pi}{2} + 2m\pi \implies \alpha = \frac{\pi}{64} + \frac{m\pi}{16} \]

### Step 9: Find a Specific Value of \(\alpha\)
Choosing \(m = 0\):
\[ \alpha = \frac{\pi}{64} \]

Thus, a value of \(\alpha\) is:
\[ \alpha = \frac{\pi}{64} \]