Question

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

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Answer 1

Answer:

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Answer 2

$❥Answer$

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.

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