In `aDeltaABC,2acsin((A-B+C)/(2))` is equal to

Options

A. `a^(2)+b^(2)-c^(2)`
B. `c^(2)+a^(2)-b^(2)`
C. `b^(2)-c^(2)-a^(2)`
D. `c^(2)-a^(2)-b^(2)`

Answer

To solve the problem, we need to find the expression for \( 2ac \sin\left(\frac{A - B + C}{2}\right) \) in triangle \( ABC \).

### Step-by-Step Solution:

1. **Understanding the Angles in Triangle**:
In triangle \( ABC \), the sum of the angles is \( 180^\circ \). Therefore, we can express \( A + B + C = 180^\circ \).

2. **Rearranging the Angles**:
From the equation \( A + B + C = 180^\circ \), we can express \( A + C \) as:
\[ A + C = 180^\circ - B \]

3. **Substituting into the Expression**:
Now, we substitute \( A + C \) into the sine function:
\[ 2ac \sin\left(\frac{A - B + C}{2}\right) = 2ac \sin\left(\frac{(A + C) - B}{2}\right) \]
This simplifies to:
\[ 2ac \sin\left(\frac{(180^\circ - B) - B}{2}\right) = 2ac \sin\left(\frac{180^\circ - 2B}{2}\right) \]

4. **Using the Sine Identity**:
We know that \( \sin(180^\circ - x) = \sin x \). Therefore:
\[ \sin\left(\frac{180^\circ - 2B}{2}\right) = \sin(90^\circ - B) = \cos B \]

5. **Final Expression**:
Substituting back, we have:
\[ 2ac \sin\left(\frac{A - B + C}{2}\right) = 2ac \cos B \]

6. **Applying the Cosine Rule**:
By the cosine rule, we know:
\[ a^2 = b^2 + c^2 - 2bc \cos A \]
Rearranging gives us:
\[ 2ac \cos B = a^2 + c^2 - b^2 \]

7. **Conclusion**:
Thus, we conclude that:
\[ 2ac \sin\left(\frac{A - B + C}{2}\right) = a^2 + c^2 - b^2 \]

### Final Answer:
The expression \( 2ac \sin\left(\frac{A - B + C}{2}\right) \) is equal to \( a^2 + c^2 - b^2 \).

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