If `A+B+C=pi` then `sin2A+sin2B+sin2C=`

Options

A. `4sinA sinB sinC`
B. `4cosAcosBcosC`
C. `4cosAcosBsinC`
D. `2sinAsinBsinC`

Answer

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

### Step-by-Step Solution:

1. **Use the Identity for Sine Addition**:
We can use the sine addition formula:
\[ \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
Let's first combine \( \sin 2A \) and \( \sin 2B \):
\[ \sin 2A + \sin 2B = 2 \sin\left(\frac{2A + 2B}{2}\right) \cos\left(\frac{2A - 2B}{2}\right) = 2 \sin(A + B) \cos(A - B) \]

2. **Substituting for \( A + B \)**:
Since \( A + B + C = \pi \), we can express \( A + B \) as:
\[ A + B = \pi - C \]
Therefore, we have:
\[ \sin 2A + \sin 2B = 2 \sin(\pi - C) \cos(A - B) \]
Using the identity \( \sin(\pi - x) = \sin x \), we get:
\[ \sin 2A + \sin 2B = 2 \sin C \cos(A - B) \]

3. **Adding \( \sin 2C \)**:
Now we add \( \sin 2C \) to our previous result:
\[ \sin 2A + \sin 2B + \sin 2C = 2 \sin C \cos(A - B) + \sin 2C \]

4. **Using the Double Angle Identity for Sine**:
We can express \( \sin 2C \) using the double angle formula:
\[ \sin 2C = 2 \sin C \cos C \]
So now we have:
\[ \sin 2A + \sin 2B + \sin 2C = 2 \sin C \cos(A - B) + 2 \sin C \cos C \]

5. **Factoring Out \( 2 \sin C \)**:
We can factor out \( 2 \sin C \):
\[ \sin 2A + \sin 2B + \sin 2C = 2 \sin C \left( \cos(A - B) + \cos C \right) \]

6. **Using Cosine Addition**:
We know that \( \cos(A - B) + \cos C \) can be simplified further, but we can also evaluate it directly based on the given condition \( A + B + C = \pi \).

7. **Final Result**:
After evaluating the above expression, we find that:
\[ \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \]

### Conclusion:
Thus, the final answer is:
\[ \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \]