Question

In any \(△ABC\), the value of \(\sin \left( \frac{A + B}{2} \right)\) is

• A. \(\sin \left( \frac{C}{2} \right)\)
• B. \(\cos \left( \frac{C}{2} \right)\)
• C. \(\sin \left( \frac{A-B}{2} \right)\)
• D. \(\cos \left( \frac{A-B}{2} \right)\)

Answer 1

The Correct Option is B

To solve the problem, we need to evaluate the expression \( \sin\left(\frac{A + B}{2}\right) \) in any triangle \( \triangle ABC \).

1. Understanding Angle Sum in a Triangle:
In any triangle, the sum of the interior angles is always:

\( A + B + C = 180^\circ \)

2. Rearranging to Express \( A + B \):
From the above identity, we get:

\( A + B = 180^\circ - C \)

3. Substituting into the Original Expression:
We are asked to find:
\( \sin\left(\frac{A + B}{2}\right) = \sin\left(\frac{180^\circ - C}{2}\right) \)

4. Applying Trigonometric Identity:
We know that:
\( \sin\left(90^\circ - x\right) = \cos(x) \)

Therefore,
\( \sin\left(\frac{180^\circ - C}{2}\right) = \cos\left(\frac{C}{2}\right) \)

Final Answer:
The value of \( \sin\left(\frac{A + B}{2}\right) \) is \( \cos\left(\frac{C}{2}\right) \)