Question:

P.T

$log[e^{i\alpha}+e^{i\alpha}]=log(2cos(\dfrac{(\alpha-\beta)}{2})+i\dfrac{(\alpha+\beta)}{2}) $

$log[e^{i\alpha}+e^{i\beta}] \\ =log[cos\alpha+isin\alpha+cos\beta+isin\beta] \\ =log(cos\alpha+cos\beta+i(sin\alpha+sin\beta)\\ =log(2 cos\dfrac{\alpha+\beta}{2}.cos\dfrac{\alpha-\beta}{2}+2isin\dfrac{\alpha+\beta}{2}.cos\dfrac{\alpha-\beta}{2} )$

$ =log(2 cos\dfrac{\alpha-\beta}{2}.[sin\dfrac{\alpha+\beta}{2}+icos\dfrac{\alpha+\beta}{2} ] \\ =log(2 cos\dfrac{\alpha-\beta}{2})+log.[cos\dfrac{\alpha+\beta}{2}+isin\dfrac{\alpha+\beta}{2} ] \\ $

$log(2 cos\dfrac{\alpha-\beta}{2})+log.[ ie^{(\dfrac{\alpha+\beta}{2}) }] \\ log(2 cos\dfrac{\alpha-\beta}{2})+i( \dfrac{\alpha+\beta}{2}).\ loge ] .since\ loge\ is 1 \\ log(2 cos\dfrac{\alpha-\beta}{2})+i( \dfrac{\alpha+\beta}{2}). Proved \\ $