if A,B,c are interior angles of triangle ABC then show that sin (B+C/2) = cos A/2

A + B + C = 180^o

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 ) = (B + C) / 2

Therefore,

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

Hence proved.

A+B+C=180

So,

b+c=180-A

b+c/2=90-a/2 ( Dividing the L.H.S and R.H.S by 2)

sin(b+c/2)

= sin ( 90 -a/2) ( Substituting b+c/2 value)

= cos a/2

Hence Proved

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