Given that `F(x)=[{:(,cos x,-sin x,0),(,sin x,cos x,0),(,0,0,1):}]."If"in R` Then for what values of y, `F(x+y)=F(x)F(y)`

Answer

To solve the problem, we need to show that \( F(x+y) = F(x)F(y) \) for certain values of \( y \). Let's break down the solution step by step.

### Step 1: Define the function \( F(x) \)
Given:
\[ F(x) = \begin{pmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Find \( F(y) \)
To find \( F(y) \), we replace \( x \) with \( y \):
\[ F(y) = \begin{pmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Find \( F(x+y) \)
Next, we compute \( F(x+y) \) by replacing \( x \) with \( x+y \):
\[ F(x+y) = \begin{pmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Compute \( F(x)F(y) \)
Now, we need to multiply \( F(x) \) and \( F(y) \):
\[ F(x)F(y) = \begin{pmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Perform the matrix multiplication
Calculating the product:
1. The first row, first column:
\[ \cos x \cos y + (-\sin x)(\sin y) = \cos x \cos y - \sin x \sin y \]
2. The first row, second column:
\[ \cos x (-\sin y) + (-\sin x)(\cos y) = -\cos x \sin y - \sin x \cos y \]
3. The second row, first column:
\[ \sin x \cos y + \cos x \sin y = \sin x \cos y + \cos x \sin y \]
4. The second row, second column:
\[ \sin x (-\sin y) + \cos x \cos y = -\sin x \sin y + \cos x \cos y \]
5. The third row remains:
\[ \begin{pmatrix} 0 & 0 & 1 \end{pmatrix} \]

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

### Step 6: Compare \( F(x+y) \) and \( F(x)F(y) \)
Using the angle addition formulas:
- \( \cos(x+y) = \cos x \cos y - \sin x \sin y \)
- \( \sin(x+y) = \sin x \cos y + \cos x \sin y \)

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

### Conclusion
Since \( F(x+y) = F(x)F(y) \) holds true for all \( x \) and \( y \), we conclude that the equation is valid for all \( y \in \mathbb{R} \).