Question

if abc are interior angles of triangle abc then show that sin B+C/2=cos A/2​

Answer 1

 Sin[(B+C)/2]

• [Since A+B+C=180 for interior angles of triangle ABC.]

then B+C=180-A.

NOW   Sin [(180-A)/2]

              =Sin[90-(A/2)]    

•  since Sin(90-A)=CosA

               =Cos(A/2)

Answer 2

Answer:

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