If f(x) = `[(cos x , - sinx,0),(sinx,cosx,0),(0,0,1)]` then show f(x) . f(y) = f(x+y)

Explanation

To show that \( f(x) \cdot f(y) = f(x+y) \) for the function defined as:

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

we will follow these steps:

### Step 1: Write down \( f(x) \) and \( f(y) \)

We have:
\[ f(x) = \begin{pmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
\[ f(y) = \begin{pmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Calculate \( f(x) \cdot f(y) \)

Now, we will multiply the two matrices \( f(x) \) and \( f(y) \):

\[ f(x) \cdot f(y) = \begin{pmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Perform the matrix multiplication

Calculating the elements of the resulting matrix:

1. First row, first column:
\[ \cos x \cdot \cos y + (-\sin x) \cdot \sin y + 0 \cdot 0 = \cos x \cos y - \sin x \sin y \]

2. First row, second column:
\[ \cos x \cdot (-\sin y) + (-\sin x) \cdot \cos y + 0 \cdot 0 = -\cos x \sin y - \sin x \cos y \]

3. First row, third column:
\[ \cos x \cdot 0 + (-\sin x) \cdot 0 + 0 \cdot 1 = 0 \]

4. Second row, first column:
\[ \sin x \cdot \cos y + \cos x \cdot \sin y + 0 \cdot 0 = \sin x \cos y + \cos x \sin y \]

5. Second row, second column:
\[ \sin x \cdot (-\sin y) + \cos x \cdot \cos y + 0 \cdot 0 = -\sin x \sin y + \cos x \cos y \]

6. Second row, third column:
\[ \sin x \cdot 0 + \cos x \cdot 0 + 0 \cdot 1 = 0 \]

7. Third row, first column:
\[ 0 \cdot \cos y + 0 \cdot \sin y + 1 \cdot 0 = 0 \]

8. Third row, second column:
\[ 0 \cdot (-\sin y) + 0 \cdot \cos y + 1 \cdot 0 = 0 \]

9. Third row, third column:
\[ 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 = 1 \]

Thus, we have:
\[ f(x) \cdot f(y) = \begin{pmatrix} \cos x \cos y - \sin x \sin y & -(\cos x \sin y + \sin x \cos y) & 0 \\ \sin x \cos y + \cos x \sin y & \cos x \cos y - \sin x \sin y & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Recognize the formulas

Using the trigonometric identities:
- \( \cos(x+y) = \cos x \cos y - \sin x \sin y \)
- \( \sin(x+y) = \sin x \cos y + \cos x \sin y \)

We can rewrite the resulting matrix as:
\[ f(x) \cdot f(y) = \begin{pmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Write down \( f(x+y) \)

Now, we compute \( f(x+y) \):
\[ f(x+y) = \begin{pmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 6: Conclusion

Since we have shown that:
\[ f(x) \cdot f(y) = f(x+y) \]
Thus, we conclude that \( f(x) \cdot f(y) = f(x+y) \).