Question

If \( F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \), prove that \(F(x)F(y) = F(x+y)\).

Hint:

The matrix \(F(x)\) is a 2D rotation matrix embedded in 3D space (using homogeneous coordinates). This property, \(F(x)F(y) = F(x+y)\), is a fundamental characteristic of rotation matrices. It signifies that performing a rotation by an angle \(y\) followed by a rotation by an angle \(x\) is equivalent to a single rotation by the angle \((x+y)\).

Answer 1

Step 1: Understanding the Concept:
This question requires us to prove a property of a function that returns a matrix. We need to perform matrix multiplication on the left-hand side (LHS) and show that the result is identical to the matrix function evaluated at \(x+y\), which is the right-hand side (RHS).
Step 2: Key Formula or Approach:
The proof relies on two main components:
1. The rule for matrix multiplication of two 3x3 matrices.
2. The trigonometric sum identities:
- \(\cos(x+y) = \cos x \cos y - \sin x \sin y\)
- \(\sin(x+y) = \sin x \cos y + \cos x \sin y\)
Step 3: Detailed Explanation:
We are given the matrix function:
\[ F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
The left-hand side (LHS) of the equation to be proved is \(F(x)F(y)\).
First, let's write out \(F(y)\):
\[ F(y) = \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Now, we perform the matrix multiplication \(F(x)F(y)\):
\[ F(x)F(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Let's compute each element of the resulting matrix:
- Row 1, Column 1: \((\cos x)(\cos y) + (-\sin x)(\sin y) + (0)(0) = \cos x \cos y - \sin x \sin y = \cos(x+y)\)
- Row 1, Column 2: \((\cos x)(-\sin y) + (-\sin x)(\cos y) + (0)(0) = -(\cos x \sin y + \sin x \cos y) = -\sin(x+y)\)
- Row 1, Column 3: \((\cos x)(0) + (-\sin x)(0) + (0)(1) = 0\)
- Row 2, Column 1: \((\sin x)(\cos y) + (\cos x)(\sin y) + (0)(0) = \sin x \cos y + \cos x \sin y = \sin(x+y)\)
- Row 2, Column 2: \((\sin x)(-\sin y) + (\cos x)(\cos y) + (0)(0) = \cos x \cos y - \sin x \sin y = \cos(x+y)\)
- Row 2, Column 3: \((\sin x)(0) + (\cos x)(0) + (0)(1) = 0\)
- Row 3, Column 1: \((0)(\cos y) + (0)(\sin y) + (1)(0) = 0\)
- Row 3, Column 2: \((0)(-\sin y) + (0)(\cos y) + (1)(0) = 0\)
- Row 3, Column 3: \((0)(0) + (0)(0) + (1)(1) = 1\)
So, the resulting matrix for the LHS is:
\[ F(x)F(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Now, let's evaluate the right-hand side (RHS), which is \(F(x+y)\). We simply replace \(x\) with \((x+y)\) in the definition of \(F(x)\):
\[ F(x+y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Comparing the results, we see that LHS = RHS.
Step 4: Final Answer:
We have shown that \(F(x)F(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}\), which is equal to \(F(x+y)\). Therefore, the identity \(F(x)F(y) = F(x+y)\) is proved.