Question

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

Answer 1

hello kalsjjsksjwjwjssss

Answer 2

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

Explanation:

In triangle ABC, sum of angles = 180 degrees.

A + B + C = 180

B + C = 180 - A

Multiplying both sides by 1/2, we get:

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

Taking Sin on both sides, we get:

Sin (B+C / 2) = Sin (90 - A/2) ------ (1)

Sin (90 -teta) = Cos teta.

Therefore, equation (1) becomes:

Sin (B+C / 2) = Sin (90 - A/2)  

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

Hence proved.