Question

If A, B and C are interior angles of triangle ABC, then show that

Sin(B+C/2)=cosA/2

Answer 1

From Angle Sum Property of Triangle,

A + B + C = 180°

→ B + C = 180 - A

Dividing both sides by 2,

→ (B + C)/2 = 90 - A/2

Multiplying both sides by sin function,

→ sin(B + C)/2 = sin(90 - A/2)

Since,

sin(90 - ∅) = cos∅

→ sin(B + C)/2 = cos A/2

Hence,Proved

Answer 2

$\huge\bf{Answer:-}$

Refer the attachment.