If `A+B+C=pi`, then `sin2A+sin2B-sin2C` is equal to

Options

A. 4sinAsinBsinC
B. 4cosAcosBsinC
C. 4cosAcosBcosC
D. None of these

Explanation

To solve the problem where \( A + B + C = \pi \) and we need to find the value of \( \sin 2A + \sin 2B - \sin 2C \), we can follow these steps:

### Step 1: Use the identity for sine addition
We know that:
\[ \sin 2A + \sin 2B = 2 \sin\left(A + B\right) \cos\left(A - B\right) \]
Since \( A + B + C = \pi \), we can express \( A + B \) as:
\[ A + B = \pi - C \]

### Step 2: Substitute \( A + B \) into the sine addition formula
Using the identity from Step 1:
\[ \sin 2A + \sin 2B = 2 \sin\left(\pi - C\right) \cos\left(A - B\right) \]
Using the property that \( \sin(\pi - x) = \sin x \):
\[ \sin 2A + \sin 2B = 2 \sin C \cos(A - B) \]

### Step 3: Substitute into the original expression
Now, we substitute this back into the expression we want to evaluate:
\[ \sin 2A + \sin 2B - \sin 2C = 2 \sin C \cos(A - B) - \sin 2C \]

### Step 4: Use the identity for \( \sin 2C \)
We also know that:
\[ \sin 2C = 2 \sin C \cos C \]
Thus, we can rewrite the expression:
\[ \sin 2A + \sin 2B - \sin 2C = 2 \sin C \cos(A - B) - 2 \sin C \cos C \]

### Step 5: Factor out \( 2 \sin C \)
Factoring out \( 2 \sin C \):
\[ \sin 2A + \sin 2B - \sin 2C = 2 \sin C \left( \cos(A - B) - \cos C \right) \]

### Step 6: Use the cosine subtraction formula
Using the cosine subtraction formula:
\[ \cos(A - B) - \cos C = -2 \sin\left(\frac{A + B + C}{2}\right) \sin\left(\frac{A + B - C}{2}\right) \]
Since \( A + B + C = \pi \), we have:
\[ \frac{A + B + C}{2} = \frac{\pi}{2} \]
Thus:
\[ \sin\left(\frac{A + B + C}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]
So:
\[ \cos(A - B) - \cos C = -2 \sin\left(\frac{A + B - C}{2}\right) \]

### Final Expression
Substituting this back, we get:
\[ \sin 2A + \sin 2B - \sin 2C = 2 \sin C \left(-2 \sin\left(\frac{A + B - C}{2}\right)\right) \]
This simplifies to:
\[ \sin 2A + \sin 2B - \sin 2C = -4 \sin C \sin\left(\frac{A + B - C}{2}\right) \]
However, since \( A + B - C = \pi - 2C \), we can conclude that:
\[ \sin 2A + \sin 2B - \sin 2C = 4 \sin C \cos A \cos B \]

### Final Answer
Thus, the final answer is:
\[ \sin 2A + \sin 2B - \sin 2C = 4 \sin C \cos A \cos B \]