`A_((alpha))=[{:(cos alpha,-sin alpha),(sin alpha,cos alpha):}],A_((beta))=[{:(cos beta,-sin beta),(sin beta,cos beta):}]`
Which one of the following is correct ?

Options

A. `A_((-alpha))A_((-beta))=A_((alpha+beta))`
B. `A_((-alpha))A_((beta))=A_((alpha-beta))`
C. `A_((alpha))A_((-beta))=A_({-(beta-alpha)})`
D. `A_((alpha))A_((beta))=A_((alpha+beta))`

Explanation

To solve the problem, we need to analyze the given matrices \( A(\alpha) \) and \( A(\beta) \) and find the relationship between their product and another matrix.

Given:
\[ A(\alpha) = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]
\[ A(\beta) = \begin{pmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{pmatrix} \]

### Step 1: Multiply the matrices \( A(\alpha) \) and \( A(\beta) \)

To find \( A(\alpha) \cdot A(\beta) \), we will multiply the two matrices:

\[ A(\alpha) \cdot A(\beta) = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \cdot \begin{pmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{pmatrix} \]

### Step 2: Calculate the elements of the resulting matrix

1. **First row, first column:**
\[ \cos \alpha \cdot \cos \beta + (-\sin \alpha) \cdot \sin \beta = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta) \]

2. **First row, second column:**
\[ \cos \alpha \cdot (-\sin \beta) + (-\sin \alpha) \cdot \cos \beta = -\cos \alpha \sin \beta - \sin \alpha \cos \beta = -(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = -\sin(\alpha + \beta) \]

3. **Second row, first column:**
\[ \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin(\alpha + \beta) \]

4. **Second row, second column:**
\[ \sin \alpha \cdot (-\sin \beta) + \cos \alpha \cdot \cos \beta = -\sin \alpha \sin \beta + \cos \alpha \cos \beta = \cos(\alpha + \beta) \]

### Step 3: Write the resulting matrix

Thus, the product \( A(\alpha) \cdot A(\beta) \) is:
\[ A(\alpha) \cdot A(\beta) = \begin{pmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) \end{pmatrix} \]

### Step 4: Identify the resulting matrix

This resulting matrix can be recognized as \( A(\alpha + \beta) \):
\[ A(\alpha + \beta) = \begin{pmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) \end{pmatrix} \]

### Conclusion

Thus, we have shown that:
\[ A(\alpha) \cdot A(\beta) = A(\alpha + \beta) \]

### Final Answer

The correct option is:
\[ A(\alpha) \cdot A(\beta) = A(\alpha + \beta) \]

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