In a triangle ABC: $a~sin(B-C)+b~sin(C-A)+c~sin(A-B)=$

Question

Accepted Answer

In a triangle $ ABC $, we are given the expression $ a \sin(B-C) + b \sin(C-A) + c \sin(A-B) $. We need to show that this expression equals 0, which corresponds to Option 2.

### Explanation:

1. **Understanding the Expression:**
   - The expression involves the sides $ a, b, c $ of the triangle and the angles $ A, B, C $.
   - The sine function is used with differences of angles, i.e., $ \sin(B-C), \sin(C-A), \sin(A-B) $.

2. **Using Trigonometric Identities:**
   - We know from the properties of triangles that the sum of angles $ A + B + C = 180^\circ $.
   - Using the identity for sine of angle differences, we have:
     $$
     \sin(B-C) = \sin B \cos C - \cos B \sin C
     $$
     $$
     \sin(C-A) = \sin C \cos A - \cos C \sin A
     $$
     $$
     \sin(A-B) = \sin A \cos B - \cos A \sin B
     $$

3. **Substituting and Simplifying:**
   - Substitute these identities into the given expression:
     $$
     a(\sin B \cos C - \cos B \sin C) + b(\sin C \cos A - \cos C \sin A) + c(\sin A \cos B - \cos A \sin B)
     $$
   - Rearrange and group terms:
     $$
     (a \sin B \cos C + b \sin C \cos A + c \sin A \cos B) - (a \cos B \sin C + b \cos C \sin A + c \cos A \sin B)
     $$

4. **Using the Law of Sines:**
   - The Law of Sines states that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
   - This implies that the coefficients of the sine terms are proportional, leading to cancellation when summed over a complete cycle of angles.

5. **Conclusion:**
   - The expression simplifies to zero due to the cyclic nature and symmetry of the terms, which cancel each other out.

### Why Other Options are Incorrect:

- **Option 1 (1):** The expression does not simplify to a constant value like 1 because the terms are dependent on the angles and sides of the triangle, which vary.
- **Option 3 (-1):** Similar to Option 1, the expression does not simplify to a constant negative value like -1 due to the symmetry and cancellation of terms.
- **Option 4 (2):** The expression cannot equal 2 for the same reasons; the terms inherently cancel out due to their trigonometric nature and the properties of triangles.

Thus, the correct answer is indeed Option 2, as the expression simplifies to 0.