If A, B and C are interior angles of a triangle ABC, then show that
sin (B+C/2) = cos A/2
In a triangle, sum of all the interior angles
A + B + C = 180°
⇒ B + C = 180° - A
⇒ (B+C)/2 = (180°-A)/2
⇒ (B+C)/2 = (90°-A/2)
⇒ sin (B+C)/2 = sin (90°-A/2)
⇒ sin (B+C)/2 = cos A/2
