If cosA+cosB=cosC,sinA+sinB=sinC
then the value of expression sin(A+B)sin2C is
Open in App
Solution
eiA+eiB=cosA+cosB+i(sinA+sinB)=eiC&e−iA+e−iB=e−iC⇒eiCeiA×eiB=e−iC⇒e2iC=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