show that if a,b and c are the interior angles of a triangle,then show that sin(b+c/2) =sin(a/2)
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a + b + c = 180 (ASP)
b + c = 180 - a
(b + c)/2 = (180 - a)/2 = 90 - a/2
sin (b+c/2) = sin(90 - a/2) = cos(a/2)
I think that what I proved is correct so check your question please
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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)]
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