If cos A + cos B = cos C and sin A + sin B = | vedclass

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(B) Given: $\cos A + \cos B = \cos C$ and $\sin A + \sin B = \sin C$.Let $z_1 = \cos A + i \sin A = e^{iA}$,$z_2 = \cos B + i \sin B = e^{iB}$,and $z_

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(B) Given: $\cos A + \cos B = \cos C$ and $\sin A + \sin B = \sin C$.Let $z_1 = \cos A + i \sin A = e^{iA}$,$z_2 = \cos B + i \sin B = e^{iB}$,and $z_3 = \cos C + i \sin C = e^{iC}$.Then $z_1 + z_2 = \cos C + i \sin C = z_3$.Also,taking the conjugate: $\cos A - i \sin A + \cos B - i \sin B = \cos C - i \sin C$,which implies $e^{-iA} + e^{-iB} = e^{-iC}$.From $e^{iA} + e^{iB} = e^{iC}$,we have $e^{iA} + e^{iB} = e^{iC}$.Dividing the two equations: $\frac{e^{iA} + e^{iB}}{e^{-iA} + e^{-iB}} = \frac{e^{iC}}{e^{-iC}} = e^{2iC}$.Since $\frac{e^{iA} + e^{iB}}{e^{-iA} + e^{-iB}} = \frac{e^{iA} + e^{iB}}{\frac{1}{e^{iA}} + \frac{1}{e^{iB}}} = \frac{e^{iA} + e^{iB}}{\frac{e^{iA} + e^{iB}}{e^{iA}e^{iB}}} = e^{i(A+B)}$.Thus,$e^{i(A+B)} = e^{2iC}$.Comparing the imaginary parts,$\sin(A+B) = \sin 2C$.Therefore,$\frac{\sin(A+B)}{\sin 2C} = 1$.

If $\cos A + \cos B = \cos C$ and $\sin A + \sin B = \sin C$,then the value of the expression $\frac{\sin(A + B)}{\sin 2C}$ is:

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