If `A+B+C=pi`, then `cos2A+cos2B+cos2C=?`

Options

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

Answer

To solve the problem where \( A + B + C = \pi \) (or 180 degrees), 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**:
Since \( A + B + C = \pi \), we can express \( C \) as \( C = \pi - A - B \).

2. **Substitute \( C \) into the cosine terms**:
We can rewrite \( \cos 2C \) using the identity:
\[ \cos 2C = \cos(2(\pi - A - B)) = \cos(2\pi - 2A - 2B) = \cos(2A + 2B) \]
(Since \( \cos(2\pi - x) = \cos x \)).

3. **Combine the cosine terms**:
Now we can rewrite the expression:
\[ \cos 2A + \cos 2B + \cos 2C = \cos 2A + \cos 2B + \cos(2A + 2B) \]

4. **Use the cosine addition formula**:
We can use the cosine addition formula:
\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
Let \( x = 2A \) and \( y = 2B \):
\[ \cos 2A + \cos 2B = 2 \cos\left(A + B\right) \cos\left(A - B\right) \]

5. **Substituting back**:
Now substituting back into our expression:
\[ \cos 2A + \cos 2B + \cos(2A + 2B) = 2 \cos(A + B) \cos(A - B) + \cos(2A + 2B) \]
Since \( A + B = \pi - C \):
\[ \cos(A + B) = \cos(\pi - C) = -\cos C \]

6. **Final expression**:
Thus, we have:
\[ \cos 2A + \cos 2B + \cos 2C = 2(-\cos C) \cos(A - B) + \cos(2A + 2B) \]
We can simplify it further, but we can also evaluate it directly:
\[ \cos 2A + \cos 2B + \cos 2C = -1 \]

### Conclusion:
The final result is:
\[ \cos 2A + \cos 2B + \cos 2C = -1 \]