show analytically that if (m) is a line through the origin with inclination angle (phi) from the positive (x)-axis and if (n) is a line through the origin with inclination angle ( heta) from the positive (x)-axis, then (r_n r_m) is a rotation about the origin by angle (2( heta - phi)).

Question

show analytically that if (m) is a line through the origin with inclination angle (phi) from the positive (x)-axis and if (n) is a line through the origin with inclination angle (	heta) from the positive (x)-axis, then (r_n r_m) is a rotation about the origin by angle (2(	heta - phi)).

Accepted Answer

# Explanation:

⚡ Using what you learned: Compositions of Transformations

## Step 1: Define the reflection matrices

A reflection across a line through the origin with inclination angle $\alpha$ is represented by the transformation matrix:
$$
R_{\alpha} = \begin{pmatrix} \cos(2\alpha) & \sin(2\alpha) \ \sin(2\alpha) & -\cos(2\alpha) \end{pmatrix}
$$

For the line $m$ with inclination angle $\phi$, the reflection matrix $r_m$ is:
$$
r_m = \begin{pmatrix} \cos(2\phi) & \sin(2\phi) \ \sin(2\phi) & -\cos(2\phi) \end{pmatrix}
$$

For the line $n$ with inclination angle $\theta$, the reflection matrix $r_n$ is:
$$
r_n = \begin{pmatrix} \cos(2\theta) & \sin(2\theta) \ \sin(2\theta) & -\cos(2\theta) \end{pmatrix}
$$

## Step 2: Compute the composition $r_n r_m$

The composition $r_n r_m$ corresponds to the matrix multiplication $r_n \cdot r_m$:
$$
r_n r_m = \begin{pmatrix} \cos(2\theta) & \sin(2\theta) \ \sin(2\theta) & -\cos(2\theta) \end{pmatrix} \begin{pmatrix} \cos(2\phi) & \sin(2\phi) \ \sin(2\phi) & -\cos(2\phi) \end{pmatrix}
$$

Multiplying the matrices yields:
$$
r_n r_m = \begin{pmatrix} \cos(2\theta)\cos(2\phi) + \sin(2\theta)\sin(2\phi) & \cos(2\theta)\sin(2\phi) - \sin(2\theta)\cos(2\phi) \ \sin(2\theta)\cos(2\phi) - \cos(2\theta)\sin(2\phi) & \sin(2\theta)\sin(2\phi) + \cos(2\theta)\cos(2\phi) \end{pmatrix}
$$

## Step 3: Simplify using trigonometric identities

Apply the angle difference identities:
* $\cos(A - B) = \cos A \cos B + \sin A \sin B$
* $\sin(A - B) = \sin A \cos B - \cos A \sin B$

Substituting $A = 2\theta$ and $B = 2\phi$:
* Entry $(1,1)$: $\cos(2\theta)\cos(2\phi) + \sin(2\theta)\sin(2\phi) = \cos(2\theta - 2\phi)$
* Entry $(1,2)$: $\cos(2\theta)\sin(2\phi) - \sin(2\theta)\cos(2\phi) = -\sin(2\theta - 2\phi)$
* Entry $(2,1)$: $\sin(2\theta)\cos(2\phi) - \cos(2\theta)\sin(2\phi) = \sin(2\theta - 2\phi)$
* Entry $(2,2)$: $\sin(2\theta)\sin(2\phi) + \cos(2\theta)\cos(2\phi) = \cos(2\theta - 2\phi)$

This gives:
$$
r_n r_m = \begin{pmatrix} \cos(2(\theta - \phi)) & -\sin(2(\theta - \phi)) \ \sin(2(\theta - \phi)) & \cos(2(\theta - \phi)) \end{pmatrix}
$$

This matrix matches the standard rotation matrix $R_{\psi}$ by an angle $\psi = 2(\theta - \phi)$ about the origin:
$$
R_{\psi} = \begin{pmatrix} \cos\psi & -\sin\psi \ \sin\psi & \cos\psi \end{pmatrix}
$$

# Answer:

By representing the reflections $r_m$ and $r_n$ as matrices:
$$
r_m = \begin{pmatrix} \cos(2\phi) & \sin(2\phi) \ \sin(2\phi) & -\cos(2\phi) \end{pmatrix}, \quad r_n = \begin{pmatrix} \cos(2\theta) & \sin(2\theta) \ \sin(2\theta) & -\cos(2\theta) \end{pmatrix}
$$
their composition $r_n r_m$ is computed via matrix multiplication:
$$
r_n r_m = \begin{pmatrix} \cos(2\theta)\cos(2\phi) + \sin(2\theta)\sin(2\phi) & \cos(2\theta)\sin(2\phi) - \sin(2\theta)\cos(2\phi) \ \sin(2\theta)\cos(2\phi) - \cos(2\theta)\sin(2\phi) & \sin(2\theta)\sin(2\phi) + \cos(2\theta)\cos(2\phi) \end{pmatrix}
$$
Applying the trigonometric angle difference identities simplifies this to:
$$
r_n r_m = \begin{pmatrix} \cos(2(\theta - \phi)) & -\sin(2(\theta - \phi)) \ \sin(2(\theta - \phi)) & \cos(2(\theta - \phi)) \end{pmatrix}
$$
which is the matrix representing a rotation about the origin by the angle $2(\theta - \phi)$.

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QUESTION IMAGE

Question

Explanation:

Step 1: Define the reflection matrices

Step 2: Compute the composition \(r_n r_m\)

Step 3: Simplify using trigonometric identities

Answer: