Question

if A,B,C are interior angles of triangle ABC prove that Cos(A + B)/2 = Sin∅/2

Answer 1

Answer:

A+B+C=180

B+C=180-A

B+C/2=90-A/2

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

Sin(B+C/2)=CosA/2A + 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 ]  

Step-by-step explanation:

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Answer 2

Answer:

A + B + C = 180°

=> (A+B+C)/2 = 90°

=> (A + B)/2 + C/2 = 90°

=> (A + B)/3 = 90° - C/2

apply cos on both sides

=> cos (A+ B)/2 = cos (90° - C/2)

=> cos (A + B) = sin C/2

[ cos (90 - x) = sin x ]

hence proved