Question

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

Answer 1

$e^{iA}+e^{iB}=\cos A+\cos B+i(\sin A+\sin B)=e^{iC}\&e^{-iA}+e^{-iB}=e^{-iC}\Rightarrow \frac{e^{iC}}{e^{iA}\times e^{iB}}=e^{-iC}\Rightarrow e^{2iC}=e^{i(A+B)}\Rightarrow \cos 2C+i\sin 2C=\cos (A+B)+i\sin (A+B)$
Comparing the imaginary part of the equation
$\sin 2C=\sin (A+B)$
$\Rightarrow \frac{\sin (A+B)}{\sin 2C}=1$