Question

If A, B, C are angles of a triangle ABC,
show Sin (B+C/2) = cos A/2.
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Answer 1

A + B + C = 180°

Therefore ,

B + C = 180 - A

(B + C) / 2 = (180 - A ) / 2

(B + C) / 2 = 90 - (A / 2)

Hence R.H.S. :- cos (A / 2) = sin [ 90 - (A / 2) ]

= sin [ (B+C) / 2 ] ----- As 90 - (A / 2) = 2 (B+C) / 2 ]

Therefore ,

cos (A / 2) = sin [ (B+C) / 2 ]

Hence proved.

Answer 2

Given:- ABC is a Triangle.

To Prove that:- Sin(B+C)/2 = Cos A/2

We know that,

=> A + B + C = 180° ( Angle Property of Triangle).

=> B + C = 180°- A

Dividing both sides by 2.

=> (B + C)/2 = 180°/2 - A/2

=> (B + C)/2 = 90° - A/2

Multiplying both sides by Sin

=> Sin(B+C)/2 = Sin (90°- A/2)

As there is a Formula

=> Sin(90°-Ø) = CosØ

=> Sin(B+C)/2 = Cos A/2

Hence, Proved