Question

If \( A = \begin{bmatrix} 0 & -\tan{\frac{\alpha}{2}} \\[4pt] \tan{\frac{\alpha}{2}} & 0 \end{bmatrix} \), then prove that \[ (I + A) = (I - A)\begin{bmatrix} \cos{\alpha} & -\sin{\alpha} \\[4pt] \sin{\alpha} & \cos{\alpha} \end{bmatrix}. \]

Hint:

For matrix problems involving trigonometric functions, look for opportunities to use trigonometric identities. The half-angle tangent identities (Weierstrass substitution) are particularly powerful for simplifying expressions involving \( \sin{\alpha} \) and \( \cos{\alpha} \).

Answer 1

Step 1: Understanding the Concept:
This problem requires proving a matrix equality. The strategy is to compute the left-hand side (LHS) and the right-hand side (RHS) of the equation separately and then show that they are equal. The proof will heavily rely on trigonometric half-angle identities.
Step 2: Key Formula or Approach:
We will use the half-angle identities for sine and cosine in terms of tangent:
\[ \sin{\alpha} = \frac{2\tan(\alpha/2)}{1 + \tan^2(\alpha/2)} \] \[ \cos{\alpha} = \frac{1 - \tan^2(\alpha/2)}{1 + \tan^2(\alpha/2)} \] Let \( t = \tan(\alpha/2) \). Then \( \sin{\alpha} = \frac{2t}{1+t^2} \) and \( \cos{\alpha} = \frac{1-t^2}{1+t^2} \).
Step 3: Detailed Explanation or Calculation:
Let \( t = \tan(\alpha/2) \). The matrix \(A\) is \[ A = \begin{bmatrix} 0 & -t \\[4pt] t & 0 \end{bmatrix}. \]
Calculate the LHS:
\[ I + A = \begin{bmatrix} 1 & 0 \\[4pt] 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -t \\[4pt] t & 0 \end{bmatrix} = \begin{bmatrix} 1 & -t \\[4pt] t & 1 \end{bmatrix}. \]
Calculate the RHS:
First, find \(I - A\):
\[ I - A = \begin{bmatrix} 1 & 0 \\[4pt] 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -t \\[4pt] t & 0 \end{bmatrix} = \begin{bmatrix} 1 & t \\[4pt] -t & 1 \end{bmatrix}. \] The rotation matrix is \[ R = \begin{bmatrix} \cos{\alpha} & -\sin{\alpha} \\[4pt] \sin{\alpha} & \cos{\alpha} \end{bmatrix}. \] Now compute the product \( (I - A)R \):
\[ \text{RHS} = \begin{bmatrix} 1 & t \\[4pt] -t & 1 \end{bmatrix} \begin{bmatrix} \cos{\alpha} & -\sin{\alpha} \\[4pt] \sin{\alpha} & \cos{\alpha} \end{bmatrix} = \begin{bmatrix} \cos{\alpha} + t\sin{\alpha} & -\sin{\alpha} + t\cos{\alpha} \\[6pt] - t\cos{\alpha} + \sin{\alpha} & t\sin{\alpha} + \cos{\alpha} \end{bmatrix}. \]
Now substitute the half-angle formulas for \( \sin{\alpha} \) and \( \cos{\alpha} \):
Entry (1,1): \[ \cos{\alpha} + t\sin{\alpha} = \frac{1-t^2}{1+t^2} + t\left(\frac{2t}{1+t^2}\right) = \frac{1-t^2+2t^2}{1+t^2} = \frac{1+t^2}{1+t^2} = 1. \]
Entry (1,2): \[ -\sin{\alpha} + t\cos{\alpha} = -\frac{2t}{1+t^2} + t\left(\frac{1-t^2}{1+t^2}\right) = \frac{-2t + t - t^3}{1+t^2} = \frac{-t - t^3}{1+t^2} = \frac{-t(1+t^2)}{1+t^2} = -t. \]
Entry (2,1): \[ - t\cos{\alpha} + \sin{\alpha} = -t\left(\frac{1-t^2}{1+t^2}\right) + \frac{2t}{1+t^2} = \frac{-t + t^3 + 2t}{1+t^2} = \frac{t + t^3}{1+t^2} = \frac{t(1+t^2)}{1+t^2} = t. \]
Entry (2,2): \[ t\sin{\alpha} + \cos{\alpha} = t\left(\frac{2t}{1+t^2}\right) + \frac{1-t^2}{1+t^2} = \frac{2t^2 + 1 - t^2}{1+t^2} = \frac{1+t^2}{1+t^2} = 1. \]
So, the RHS matrix is:
\[ \text{RHS} = \begin{bmatrix} 1 & -t \\[4pt] t & 1 \end{bmatrix}. \]
Step 4: Final Answer:
We have shown that LHS \(= \begin{bmatrix} 1 & -t \\[4pt] t & 1 \end{bmatrix}\) and RHS \(= \begin{bmatrix} 1 & -t \\[4pt] t & 1 \end{bmatrix}\).
Since LHS = RHS, the identity is proved.