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Question

If cosA+cosB=cosC,sinA+sinB=sinC
then the value of expression sin(A+B)sin2C is

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Solution

eiA+eiB = cosA+cosB+i(sinA+sinB) = eiC&eiA+eiB = eiCeiCeiA×eiB = eiCe2iC = ei(A+B)cos2C+isin2C = cos(A+B)+isin(A+B)
Comparing the imaginary part of the equation
sin2C = sin(A+B)
sin(A+B)sin2C = 1

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