Question

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

Hey

Consider a,b and c to be the angles of the triangle.

Now,

Angle sum property,

a+b+c=180

=>b+c=180-a

Dividing by 2 on both sides,

(b+c)/2=90-a/2

Multiplying by sin on both sides,

=>sin[(b+c)/2]=sin(90-a/2)

=>sin[(b+c)/2]=cosa/2

Hence,proved

Answer 2

answer is In the above attachment