Question

If \(A\), \(B\), and \(C\) are interior angles of triangle \(ABC\), then the value of \(\cos \left( \frac{A + B}{2} \right)\) is:

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

Hint:

In a triangle, the sum of the interior angles is \(180^\circ\). Use this property to simplify trigonometric expressions involving the angles of a triangle.

Answer 1

The Correct Option is C

In any triangle, the sum of the interior angles is \(180^\circ\). Hence, \[ A + B + C = 180^\circ \] This implies: \[ A + B = 180^\circ - C \] Thus, we can write: \[ \cos \left( \frac{A + B}{2} \right) = \cos \left( \frac{180^\circ - C}{2} \right) = \sin \left( \frac{C}{2} \right) \] Thus, the correct answer is option (3).