SOLUTION: Let a, b, c be three sides opposite to the angles A, B, C respectively in a triangle ABC. Find 2ac.Sin(A + C - B)/2?

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Question 1184170: Let a, b, c be three sides opposite to the angles A, B, C respectively in a triangle ABC. Find 2ac.Sin(A + C - B)/2?
Answer by ikleyn(52818) About Me  (Show Source):
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Let a, b, c be three sides opposite to the angles A, B, C respectively in a triangle ABC.
Find 2ac.Sin((A + C - B)/2).
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First,  %28%28A+%2B+C+-+B%29%2F2%29 = %28%28A%2BB%2BC%29-2B%29%2F2 = %28pi-2B%29%2F2 = pi%2F2+-+B.

THEREFORE,  sin%28%28A+%2B+C+-+B%29%2F2%29 = sin%28pi%2F2-B%29 = cos(B).



Next, due to the cosine law for triangle ABC

    b%5E2 = a%5E2 + c%5E2 - 2ac%2Acos%28B%29.



It implies

    2ac*cos(B) = a^2 + c^2 - b^2



Now we can write the final identity

    2ac%2Asin%28%28A%2BC-B%29%2F2%29 = 2ac*cos(B) = a%5E2+%2B+c%5E2+-+b%5E2.



ANSWER.  2ac%2Asin%28%28A%2BC-B%29%2F2%29 = a%5E2+%2B+c%5E2+-+b%5E2.

Solved.