Let `F(alpha)=[(cos alpha, -sin alpha, 0),(sin alpha, cos alpha, 0),(0,0,1)]` then `F(alpha). F (beta)` is equal to

Options

A. `F(alpha beta)`
B. `F((alpha)/(beta))`
C. `F(alpha+beta)`
D. `F(alpha-beta)`

Answer

To solve the problem, we need to multiply the matrices \( F(\alpha) \) and \( F(\beta) \). Let's denote:

\[ F(\alpha) = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

\[ F(\beta) = \begin{pmatrix} \cos \beta & -\sin \beta & 0 \\ \sin \beta & \cos \beta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Now, we will perform the matrix multiplication \( F(\alpha) \cdot F(\beta) \).

### Step 1: Multiply the first row of \( F(\alpha) \) with each column of \( F(\beta) \)

1. **First Column:**
\[ \text{First row} \cdot \text{First column} = \cos \alpha \cdot \cos \beta + (-\sin \alpha) \cdot \sin \beta = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

2. **Second Column:**
\[ \text{First row} \cdot \text{Second column} = \cos \alpha \cdot (-\sin \beta) + (-\sin \alpha) \cdot \cos \beta = -\cos \alpha \sin \beta - \sin \alpha \cos \beta \]

3. **Third Column:**
\[ \text{First row} \cdot \text{Third column} = \cos \alpha \cdot 0 + (-\sin \alpha) \cdot 0 + 0 \cdot 1 = 0 \]

### Step 2: Multiply the second row of \( F(\alpha) \) with each column of \( F(\beta) \)

1. **First Column:**
\[ \text{Second row} \cdot \text{First column} = \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]

2. **Second Column:**
\[ \text{Second row} \cdot \text{Second column} = \sin \alpha \cdot (-\sin \beta) + \cos \alpha \cdot \cos \beta = -\sin \alpha \sin \beta + \cos \alpha \cos \beta \]

3. **Third Column:**
\[ \text{Second row} \cdot \text{Third column} = \sin \alpha \cdot 0 + \cos \alpha \cdot 0 + 0 \cdot 1 = 0 \]

### Step 3: Multiply the third row of \( F(\alpha) \) with each column of \( F(\beta) \)

1. **First Column:**
\[ \text{Third row} \cdot \text{First column} = 0 \cdot \cos \beta + 0 \cdot \sin \beta + 1 \cdot 0 = 0 \]

2. **Second Column:**
\[ \text{Third row} \cdot \text{Second column} = 0 \cdot (-\sin \beta) + 0 \cdot \cos \beta + 1 \cdot 0 = 0 \]

3. **Third Column:**
\[ \text{Third row} \cdot \text{Third column} = 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 = 1 \]

### Step 4: Combine the results

Now we can combine all the results to form the resulting matrix:

\[ F(\alpha) \cdot F(\beta) = \begin{pmatrix} \cos \alpha \cos \beta - \sin \alpha \sin \beta & -(\cos \alpha \sin \beta + \sin \alpha \cos \beta) & 0 \\ \sin \alpha \cos \beta + \cos \alpha \sin \beta & \cos \alpha \cos \beta - \sin \alpha \sin \beta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Simplify using trigonometric identities

Using the angle addition formulas:
- \( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)
- \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)

We can rewrite the matrix as:

\[ F(\alpha) \cdot F(\beta) = \begin{pmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) & 0 \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Final Result

Thus, we have:

\[ F(\alpha) \cdot F(\beta) = F(\alpha + \beta) \]