If `E(theta)=[[cos theta, sin theta] , [-sin theta, cos theta]]` then `E(alpha) E(beta)=`

Options

A. `E(0^@)`
B. `E(alphabeta)`
C. `E(alpha+beta)`
D. `E(alpha-beta)`

Answer

To solve the problem, we need to find the product of two matrices \( E(\alpha) \) and \( E(\beta) \), where:

\[ E(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \]

### Step-by-Step Solution:

1. **Find \( E(\alpha) \)**:
Substitute \( \theta \) with \( \alpha \) in the matrix \( E(\theta) \):
\[ E(\alpha) = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \]

**Hint**: Remember to replace \( \theta \) with \( \alpha \) in the matrix.

2. **Find \( E(\beta) \)**:
Substitute \( \theta \) with \( \beta \) in the matrix \( E(\theta) \):
\[ E(\beta) = \begin{bmatrix} \cos \beta & \sin \beta \\ -\sin \beta & \cos \beta \end{bmatrix} \]

**Hint**: Similarly, replace \( \theta \) with \( \beta \) in the matrix.

3. **Multiply \( E(\alpha) \) and \( E(\beta) \)**:
We need to compute the product \( E(\alpha) E(\beta) \):
\[ E(\alpha) E(\beta) = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \beta & \sin \beta \\ -\sin \beta & \cos \beta \end{bmatrix} \]

**Hint**: Use the matrix multiplication rule (row by column).

4. **Calculate the elements of the resulting matrix**:
- First element (1,1):
\[ \cos \alpha \cos \beta + \sin \alpha (-\sin \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

- Second element (1,2):
\[ \cos \alpha \sin \beta + \sin \alpha \cos \beta \]

- Third element (2,1):
\[ -\sin \alpha \cos \beta + \cos \alpha (-\sin \beta) = -\sin \alpha \cos \beta - \cos \alpha \sin \beta \]

- Fourth element (2,2):
\[ -\sin \alpha \sin \beta + \cos \alpha \cos \beta \]

Thus, the product matrix is:
\[ E(\alpha) E(\beta) = \begin{bmatrix} \cos \alpha \cos \beta - \sin \alpha \sin \beta & \cos \alpha \sin \beta + \sin \alpha \cos \beta \\ -\sin \alpha \cos \beta - \cos \alpha \sin \beta & -\sin \alpha \sin \beta + \cos \alpha \cos \beta \end{bmatrix} \]

5. **Use trigonometric identities**:
From trigonometric identities, we know:
- \( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)
- \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)

Therefore, we can rewrite the matrix as:
\[ E(\alpha) E(\beta) = \begin{bmatrix} \cos(\alpha + \beta) & \sin(\alpha + \beta) \\ -\sin(\alpha + \beta) & \cos(\alpha + \beta) \end{bmatrix} \]

6. **Conclude**:
We can see that:
\[ E(\alpha) E(\beta) = E(\alpha + \beta) \]

**Final Answer**:
\[ E(\alpha) E(\beta) = E(\alpha + \beta) \]