The value of the determinant
`Delta = |(cos (alpha + beta),- sin (alpha + beta),cos 2 beta),(sin alpha,cos alpha,sin beta),(- cos alpha,sin alpha,- cos beta)|`, is

Options

A. `cos^(2) alpha`
B. `sin^(2) alpha`
C. `sin (alpha - beta)`
D. 0

Explanation

To find the value of the determinant

\[ \Delta = \begin{vmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) & \cos(2\beta) \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & -\cos \beta \end{vmatrix} \]

we will perform row operations to simplify the determinant.

### Step 1: Apply Row Operations
We will perform the following row operations:
- Replace \( R_1 \) with \( R_1 + \sin \beta R_2 + \cos \beta R_3 \).

This gives us:

\[ R_1 = \begin{pmatrix} \cos(\alpha + \beta) + \sin \beta \sin \alpha - \cos \beta \cos \alpha & -\sin(\alpha + \beta) + \sin \beta \cos \alpha + \cos \beta \sin \alpha & \cos(2\beta) + \sin \beta \sin \beta - \cos \beta \cos \beta \end{pmatrix} \]

### Step 2: Simplify the First Row
Now we simplify the first row:

1. The first element becomes:
\[ \cos(\alpha + \beta) + \sin \beta \sin \alpha - \cos \beta \cos \alpha = \cos(\alpha + \beta) + \sin \beta \sin \alpha - \cos \beta \cos \alpha \]
Using the cosine addition formula, this simplifies to \( 0 \).

2. The second element becomes:
\[ -\sin(\alpha + \beta) + \sin \beta \cos \alpha + \cos \beta \sin \alpha = -\sin(\alpha + \beta) + \sin(\alpha + \beta) = 0 \]

3. The third element becomes:
\[ \cos(2\beta) + \sin^2 \beta - \cos^2 \beta = \cos(2\beta) + \sin^2 \beta - \cos^2 \beta = 0 \]

Thus, the first row becomes:

\[ R_1 = \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} \]

### Step 3: Evaluate the Determinant
Since the first row of the determinant is now all zeros, the value of the determinant is:

\[ \Delta = 0 \]

### Final Answer
The value of the determinant \( \Delta \) is \( 0 \).

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