If A, B, C are the angles of a triangle, then `sin 2A + sin 2B - sin 2C` is equal to

Options

A. `4 sin A cos B cos C`
B. ` 4 cos A`
C. ` 4 sin A cos A`
D. `4 cos A cos B sin C`

Answer

To solve the problem `sin 2A + sin 2B - sin 2C` where A, B, and C are the angles of a triangle, we will use some trigonometric identities and properties of angles in a triangle.

### Step-by-Step Solution:

1. **Use the property of angles in a triangle:**
Since A, B, and C are angles of a triangle, we know that:
\[ A + B + C = \pi \]
This implies:
\[ A + B = \pi - C \]

2. **Substitute A + B into the sine function:**
We can express `sin 2A + sin 2B` using the sine addition formula:
\[ \sin 2A + \sin 2B = 2 \sin(A + B) \cos(A - B) \]
Substituting \(A + B = \pi - C\):
\[ \sin 2A + \sin 2B = 2 \sin(\pi - C) \cos(A - B) \]
Since \(\sin(\pi - C) = \sin C\):
\[ \sin 2A + \sin 2B = 2 \sin C \cos(A - B) \]

3. **Combine with the third term:**
Now, we need to consider the expression:
\[ \sin 2A + \sin 2B - \sin 2C \]
We already have:
\[ \sin 2A + \sin 2B = 2 \sin C \cos(A - B) \]
Thus, we can write:
\[ \sin 2A + \sin 2B - \sin 2C = 2 \sin C \cos(A - B) - \sin 2C \]

4. **Express sin 2C:**
Using the double angle formula for sine:
\[ \sin 2C = 2 \sin C \cos C \]
Therefore, we can rewrite our expression:
\[ 2 \sin C \cos(A - B) - 2 \sin C \cos C \]

5. **Factor out common terms:**
Factoring out \(2 \sin C\):
\[ 2 \sin C (\cos(A - B) - \cos C) \]

6. **Use the cosine subtraction formula:**
We can use the identity for cosine subtraction:
\[ \cos(A - B) - \cos C = -2 \sin\left(\frac{A + B + C}{2}\right) \sin\left(\frac{A + B - C}{2}\right) \]
Since \(A + B + C = \pi\), we have:
\[ \sin\left(\frac{A + B + C}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]
Therefore:
\[ \cos(A - B) - \cos C = -2 \sin\left(\frac{C}{2}\right) \sin\left(\frac{A + B - C}{2}\right) \]

7. **Final expression:**
Thus, we can express our final answer as:
\[ \sin 2A + \sin 2B - \sin 2C = 4 \sin C \sin\left(\frac{A + B - C}{2}\right) \sin\left(\frac{C}{2}\right) \]

### Final Answer:
\[ \sin 2A + \sin 2B - \sin 2C = 4 \sin C \cos A \cos B \]