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

Options

A. `1+4cosAcosBsinC`
B. `-1+4sinAsinBcosC`
C. `-1-4cosAcosBcosC`
D. none of these

Answer

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

### Step-by-Step Solution:

1. **Use the Identity for Cosine**:
We know that \( C = \pi - (A + B) \). Therefore, we can express \( \cos 2C \) using the cosine of a sum:
\[ \cos 2C = \cos(2(\pi - (A + B))) = \cos(2\pi - 2(A + B)) = \cos(-2(A + B)) = \cos(2(A + B)) \]

2. **Apply the Cosine Addition Formula**:
Now, we can use the cosine addition formula:
\[ \cos(2(A + B)) = \cos(2A + 2B) = \cos 2A \cos 2B - \sin 2A \sin 2B \]

3. **Combine the Terms**:
Now we can express \( \cos 2A + \cos 2B + \cos 2C \):
\[ \cos 2A + \cos 2B + \cos 2C = \cos 2A + \cos 2B + \cos(2(A + B)) \]
Using the identity from step 1:
\[ = \cos 2A + \cos 2B + \cos(2A + 2B) \]

4. **Use the Cosine Sum Formula**:
We can apply the cosine sum formula again:
\[ \cos 2A + \cos 2B = 2 \cos\left(\frac{2A + 2B}{2}\right) \cos\left(\frac{2A - 2B}{2}\right) = 2 \cos(A + B) \cos(A - B) \]
Now substituting back:
\[ \cos 2A + \cos 2B + \cos(2A + 2B) = 2 \cos(A + B) \cos(A - B) + \cos(2A + 2B) \]

5. **Final Expression**:
Since \( A + B = \pi - C \):
\[ \cos(A + B) = \cos(\pi - C) = -\cos C \]
Thus, we can rewrite:
\[ \cos 2A + \cos 2B + \cos 2C = 2(-\cos C) \cos(A - B) + \cos(2A + 2B) \]

6. **Conclusion**:
After simplifying, we find:
\[ \cos 2A + \cos 2B + \cos 2C = -1 \]

### Final Answer:
\[ \cos 2A + \cos 2B + \cos 2C = -1 \]