If `Delta = |(cos (alpha_(1) - beta_(1)),cos (alpha_(1) - beta_(2)),cos (alpha_(1) - beta_(3))),(cos (alpha_(2) - beta_(1)),cos (alpha_(2) - beta_(2)),cos (alpha_(2) - beta_(3))),(cos (alpha_(3) - beta_(1)),cos (alpha_(3) - beta_(2)),cos (alpha_(3) - beta_(3)))|" then " Delta` equals

Options

A. `cos alpha_(1) cos alpha_(2) cos alpha_(3) cos beta_(1) cos beta_(2) cos beta_(3)`
B. `cos alpha_(1) + cos alpha_(2) + cos alpha_(3) + cos beta_(1) + cos beta_(2) + cos beta_(3)`
C. `cos (alpha_(1) - beta_(1)) cos (alpha_(2) - beta_(2)) cos (alpha_(3) - beta_(3))`
D. none of these

Explanation

To solve the determinant given by

\[ \Delta = \begin{vmatrix} \cos(\alpha_1 - \beta_1) & \cos(\alpha_1 - \beta_2) & \cos(\alpha_1 - \beta_3) \\ \cos(\alpha_2 - \beta_1) & \cos(\alpha_2 - \beta_2) & \cos(\alpha_2 - \beta_3) \\ \cos(\alpha_3 - \beta_1) & \cos(\alpha_3 - \beta_2) & \cos(\alpha_3 - \beta_3) \end{vmatrix} \]

we can use the trigonometric identity for cosine of a difference:

\[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]

### Step 1: Expand the determinant using the cosine difference identity

Using the identity, we can rewrite each entry of the determinant:

- For the first row:
- \(\cos(\alpha_1 - \beta_1) = \cos \alpha_1 \cos \beta_1 + \sin \alpha_1 \sin \beta_1\)
- \(\cos(\alpha_1 - \beta_2) = \cos \alpha_1 \cos \beta_2 + \sin \alpha_1 \sin \beta_2\)
- \(\cos(\alpha_1 - \beta_3) = \cos \alpha_1 \cos \beta_3 + \sin \alpha_1 \sin \beta_3\)

- For the second row:
- \(\cos(\alpha_2 - \beta_1) = \cos \alpha_2 \cos \beta_1 + \sin \alpha_2 \sin \beta_1\)
- \(\cos(\alpha_2 - \beta_2) = \cos \alpha_2 \cos \beta_2 + \sin \alpha_2 \sin \beta_2\)
- \(\cos(\alpha_2 - \beta_3) = \cos \alpha_2 \cos \beta_3 + \sin \alpha_2 \sin \beta_3\)

- For the third row:
- \(\cos(\alpha_3 - \beta_1) = \cos \alpha_3 \cos \beta_1 + \sin \alpha_3 \sin \beta_1\)
- \(\cos(\alpha_3 - \beta_2) = \cos \alpha_3 \cos \beta_2 + \sin \alpha_3 \sin \beta_2\)
- \(\cos(\alpha_3 - \beta_3) = \cos \alpha_3 \cos \beta_3 + \sin \alpha_3 \sin \beta_3\)

### Step 2: Substitute into the determinant

Now, substituting these expansions into the determinant, we have:

\[ \Delta = \begin{vmatrix} \cos \alpha_1 \cos \beta_1 + \sin \alpha_1 \sin \beta_1 & \cos \alpha_1 \cos \beta_2 + \sin \alpha_1 \sin \beta_2 & \cos \alpha_1 \cos \beta_3 + \sin \alpha_1 \sin \beta_3 \\ \cos \alpha_2 \cos \beta_1 + \sin \alpha_2 \sin \beta_1 & \cos \alpha_2 \cos \beta_2 + \sin \alpha_2 \sin \beta_2 & \cos \alpha_2 \cos \beta_3 + \sin \alpha_2 \sin \beta_3 \\ \cos \alpha_3 \cos \beta_1 + \sin \alpha_3 \sin \beta_1 & \cos \alpha_3 \cos \beta_2 + \sin \alpha_3 \sin \beta_2 & \cos \alpha_3 \cos \beta_3 + \sin \alpha_3 \sin \beta_3 \end{vmatrix} \]

### Step 3: Factor the determinant

Notice that the determinant can be expressed as a product of two separate determinants:

\[ \Delta = \begin{vmatrix} \cos \alpha_1 & \cos \alpha_2 & \cos \alpha_3 \\ \sin \alpha_1 & \sin \alpha_2 & \sin \alpha_3 \\ 0 & 0 & 0 \end{vmatrix} \cdot \begin{vmatrix} \cos \beta_1 & \cos \beta_2 & \cos \beta_3 \\ \sin \beta_1 & \sin \beta_2 & \sin \beta_3 \\ 0 & 0 & 0 \end{vmatrix} \]

### Step 4: Evaluate the determinant

Since both determinants have a row of zeros, we find that:

\[ \Delta = 0 \]

### Conclusion

Thus, the value of the determinant \(\Delta\) is:

\[ \Delta = 0 \]