If A+B+C `=(3pi)/(2)`, then `cos2A+cos2B+cos2C` is equal to-

Options

A. `1-4cosAcosBcosC`
B. `4sinAsinBsinC`
C. `1+2cosA+cosBcosC`
D. `1-4sinAsinBsinC`

Explanation

To solve the problem where \( A + B + C = \frac{3\pi}{2} \) and we need to find \( \cos 2A + \cos 2B + \cos 2C \), we can follow these steps:

### Step 1: Write the expression
We start with the expression we need to evaluate:
\[ \cos 2A + \cos 2B + \cos 2C \]

### Step 2: Use the cosine addition formula
We can use the cosine addition formula for two angles:
\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
Applying this to \( \cos 2A + \cos 2B \):
\[ \cos 2A + \cos 2B = 2 \cos\left(A + B\right) \cos\left(A - B\right) \]
So, we rewrite our expression:
\[ \cos 2A + \cos 2B + \cos 2C = 2 \cos(A + B) \cos(A - B) + \cos 2C \]

### Step 3: Substitute \( C \)
From the given condition \( A + B + C = \frac{3\pi}{2} \), we can express \( C \) as:
\[ C = \frac{3\pi}{2} - (A + B) \]
Now, we can find \( \cos 2C \):
\[ \cos 2C = \cos\left(2\left(\frac{3\pi}{2} - (A + B)\right)\right) = \cos\left(3\pi - 2(A + B)\right) \]
Using the property \( \cos(3\pi - x) = -\cos x \):
\[ \cos 2C = -\cos(2(A + B)) \]

### Step 4: Substitute back into the expression
Now we substitute back into our expression:
\[ \cos 2A + \cos 2B + \cos 2C = 2 \cos(A + B) \cos(A - B) - \cos(2(A + B)) \]

### Step 5: Use the identity for \( \cos(2x) \)
Using the identity \( \cos(2x) = 2\cos^2(x) - 1 \):
\[ \cos(2(A + B)) = 2\cos^2(A + B) - 1 \]
Thus, our expression becomes:
\[ 2 \cos(A + B) \cos(A - B) - (2\cos^2(A + B) - 1) \]
Simplifying this gives:
\[ 2 \cos(A + B) \cos(A - B) - 2\cos^2(A + B) + 1 \]

### Step 6: Factor out common terms
We can factor this expression:
\[ = 2\left(\cos(A + B) \cos(A - B) - \cos^2(A + B)\right) + 1 \]

### Step 7: Final simplification
This expression can be further simplified or evaluated based on specific values of \( A \) and \( B \) if needed. However, the general form is:
\[ \cos 2A + \cos 2B + \cos 2C = 1 - 2\sin A \sin B \]

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