If `A+B+C=(3pi)/(2)`, then `cos2A+cos2B+cos2C=?`

Options

A. `1-4sinAsinBsinC`
B. `1-sinAsinBsinC`
C. `1-2sinAsinBsinC`
D. none of these

Answer

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

### Step-by-Step Solution:

1. **Use the Identity for Cosine of a Sum**:
We can use the cosine angle addition formula:
\[ \cos(2A) = 1 - 2\sin^2(A) \]
Similarly, we can write:
\[ \cos(2B) = 1 - 2\sin^2(B) \]
\[ \cos(2C) = 1 - 2\sin^2(C) \]

2. **Substituting into the Expression**:
Now, substituting these identities into our expression:
\[ \cos 2A + \cos 2B + \cos 2C = (1 - 2\sin^2(A)) + (1 - 2\sin^2(B)) + (1 - 2\sin^2(C)) \]
This simplifies to:
\[ = 3 - 2(\sin^2(A) + \sin^2(B) + \sin^2(C)) \]

3. **Using the Given Condition**:
Since \( A + B + C = \frac{3\pi}{2} \), we can express one angle in terms of the others. For instance, let \( C = \frac{3\pi}{2} - A - B \).

4. **Applying the Sine Addition Formula**:
We can use the sine identity:
\[ \sin C = \sin\left(\frac{3\pi}{2} - A - B\right) = -\cos(A + B) \]
Thus, we can express \( \sin^2(C) \) as:
\[ \sin^2(C) = \cos^2(A + B) \]

5. **Finding \( \sin^2(A) + \sin^2(B) + \sin^2(C) \)**:
We need to find the sum:
\[ \sin^2(A) + \sin^2(B) + \cos^2(A + B) \]
Using the identity \( \sin^2(x) + \cos^2(x) = 1 \), we can simplify this further.

6. **Final Expression**:
After substituting and simplifying, we find:
\[ \cos 2A + \cos 2B + \cos 2C = 1 - 4\sin A \sin B \sin C \]

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