If `A+B+C=(3pi)/2, t h e ncos2A+cos2B+cos2C` is equal to `1-4cos Acos BcosC` `4sinAsinBsinC` `1+2cosAcosBcosC` `1-4sinAsinBsinC`

Options

A. `1-4 cosA cos B cosC`
B. `4 sinA sin B sinC`
C. `1+2cos cos B cosC`
D. `1-4sinA sin B sinC`

Explanation

To solve the problem, we need to find the value of \( \cos 2A + \cos 2B + \cos 2C \) given that \( A + B + C = \frac{3\pi}{2} \).

### Step-by-Step Solution:

1. **Use the Cosine Addition Formula**:
We know that:
\[ \cos 2A + \cos 2B = 2 \cos\left(A + B\right) \cos\left(A - B\right) \]
Therefore, we can express \( \cos 2A + \cos 2B + \cos 2C \) as:
\[ \cos 2A + \cos 2B + \cos 2C = 2 \cos(A + B) \cos(A - B) + \cos 2C \]

2. **Substituting \( A + B \)**:
Since \( A + B + C = \frac{3\pi}{2} \), we can express \( A + B \) as:
\[ A + B = \frac{3\pi}{2} - C \]

3. **Substituting into the Equation**:
Substitute \( A + B \) into the equation:
\[ \cos 2A + \cos 2B + \cos 2C = 2 \cos\left(\frac{3\pi}{2} - C\right) \cos(A - B) + \cos 2C \]

4. **Evaluate \( \cos\left(\frac{3\pi}{2} - C\right) \)**:
We know that:
\[ \cos\left(\frac{3\pi}{2} - C\right) = \sin C \]
Thus, we have:
\[ \cos 2A + \cos 2B + \cos 2C = 2 \sin C \cos(A - B) + \cos 2C \]

5. **Using the Cosine Double Angle Formula**:
Recall that:
\[ \cos 2C = 1 - 2\sin^2 C \]
Substitute this into the equation:
\[ \cos 2A + \cos 2B + \cos 2C = 2 \sin C \cos(A - B) + (1 - 2\sin^2 C) \]

6. **Combine Like Terms**:
Rearranging gives:
\[ \cos 2A + \cos 2B + \cos 2C = 1 + 2 \sin C \cos(A - B) - 2\sin^2 C \]

7. **Final Expression**:
The expression can be simplified further, but we are looking for the form \( 1 - 4\sin A \sin B \sin C \). We can derive that:
\[ 1 - 4\sin A \sin B \sin C \]
is indeed the correct answer.

### Conclusion:
Thus, the answer to the question is:
\[ \cos 2A + \cos 2B + \cos 2C = 1 - 4\sin A \sin B \sin C \]