Question

In $\triangle ABC$, with usual notations, $2ab \sin \frac{1}{2}(A+B-C) =$

• A. $a^2 - b^2 - c^2$
• B. $a^2 + b^2 - c^2$
• C. $a^2 + b^2 + c^2$
• D. $a^2 - b^2 + c^2$

Hint:

For any matching identity problem in triangles, test a simple equilateral triangle configuration where $A = B = C = 60^\circ$ and $a = b = c = 1$. $$\text{LHS} = 2(1)(1) \sin\left(\frac{60^\circ}{2}\right) = 2 \sin(30^\circ) = 1$$ Plugging $a=b=c=1$ into Option (B) yields $1^2 + 1^2 - 1^2 = 1$, revealing the answer immediately!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to simplify a trigonometric expression involving the angles and sides of a general triangle $ABC$ using standard triangle properties.

Step 2: Key Formula or Approach:
We will use the angle sum property of a triangle ($A + B + C = 180^\circ$), basic trigonometric reduction formulas, and the Cosine Rule: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$

Step 3: Detailed Explanation:
In any $\triangle ABC$, we know that: $$A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C$$ Let's substitute this identity into the argument of the sine function: $$A + B - C = (180^\circ - C) - C = 180^\circ - 2C$$ Dividing this by 2 inside the sine expression: $$\sin \frac{1}{2}(A + B - C) = \sin \left(\frac{180^\circ - 2C}{2}\right) = \sin(90^\circ - C)$$ Using the trigonometric co-function identity $\sin(90^\circ - C) = \cos C$: $$\sin \frac{1}{2}(A + B - C) = \cos C$$ Now substitute this back into the target expression: $$\text{Expression} = 2ab \cos C$$ From the standard Law of Cosines, we substitute $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$: $$\text{Expression} = 2ab \cdot \left(\frac{a^2 + b^2 - c^2}{2ab}\right)$$ The term $2ab$ cancels completely from the numerator and denominator: $$\text{Expression} = a^2 + b^2 - c^2$$

Step 4: Final Answer:
The simplified expression matches option (B).