If `A=((a_(i j)))_(nxxn)` and `f` is a function, we define `f(A)=((f(a_(i j))))_(nxxn ')` Let `A=[(pi//2-theta , theta),(-theta , pi//2-theta)]` . Then
a. `sinA` is invertible b. `sinA=cos A` c. `sinA` is orthogonal
d. `sin(2A)=2AsinA cos A`

Options

A. `sin A` is invertible
B. `sin A = cos A`
C. `sin A` is orthogonal
D. `sin (2A)=2 sin A cos A`

Explanation

To solve the problem, we need to analyze the matrix \( A \) and compute \( \sin A \) and \( \cos A \). Let's go through the steps systematically.

### Step 1: Define the Matrix A
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} \frac{\pi}{2} - \theta & \theta \\ -\theta & \frac{\pi}{2} - \theta \end{pmatrix} \]

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