Question

IF A B AND C ARE INTERIOR ANGLES OF A TRIANGLE THEN SHOW THAT
SIN[(B+C)/2]=COSA/2​

Answer 1

SEE THE AATACHMENT YOUR ANSWER IS THIS AND I THINK IT IS CORRECT

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Answer 2

Answer:

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