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Question:

If A, B, and C are interior angles of a triangle ABC, then show that sin((B+C)/2) = cos(A/2)

Solution:

∠A + ∠B + ∠C = 180° [Angle sum property of a triangle]
∠B + ∠C = 180° - ∠A
(∠B + ∠C)/2 = 90° - ∠A/2
sin((B+C)/2) = sin(90° - A/2) = cos(A/2)
Hence Proved.