If `(A+B+C)=180^(@)`, then what is `Sin2A-Sin2B-Sin2C=`

Options

A. A. `-4SinASinBSinC`
B. B. `-4CosASinBCosC`
C. C. `-4CosACosBSinC`
D. D. `-4SinACosBCosC`

Answer

To solve the problem, we need to find the value of \( \sin 2A - \sin 2B - \sin 2C \) given that \( A + B + C = 180^\circ \).

### Step-by-Step Solution:

1. **Start with the expression:**
\[ \sin 2A - \sin 2B - \sin 2C \]

2. **Rearrange the expression:**
\[ \sin 2A - (\sin 2B + \sin 2C) \]

3. **Use the sine addition formula:**
We know that:
\[ \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
Applying this to \( \sin 2B + \sin 2C \):
\[ \sin 2B + \sin 2C = 2 \sin\left(\frac{2B + 2C}{2}\right) \cos\left(\frac{2B - 2C}{2}\right) = 2 \sin(B + C) \cos(B - C) \]

4. **Substituting \( B + C \):**
Since \( A + B + C = 180^\circ \), we have:
\[ B + C = 180^\circ - A \]
Thus:
\[ \sin(B + C) = \sin(180^\circ - A) = \sin A \]
Therefore:
\[ \sin 2B + \sin 2C = 2 \sin A \cos(B - C) \]

5. **Substituting back into the expression:**
Now we substitute back into our rearranged expression:
\[ \sin 2A - 2 \sin A \cos(B - C) \]

6. **Factor out \( \sin A \):**
We can factor \( \sin A \) from the expression:
\[ = \sin A \left( \frac{\sin 2A}{\sin A} - 2 \cos(B - C) \right) \]
We know that:
\[ \frac{\sin 2A}{\sin A} = 2 \cos A \]
Thus:
\[ = \sin A \left( 2 \cos A - 2 \cos(B - C) \right) \]

7. **Final expression:**
This simplifies to:
\[ = 2 \sin A (\cos A - \cos(B - C)) \]

### Conclusion:
The final result is:
\[ \sin 2A - \sin 2B - \sin 2C = 2 \sin A (\cos A - \cos(B - C)) \]