Proving that sin⁡(z+w)=sin⁡(w)cos⁡(z)+sin⁡(z)cos⁡(w) using complex exponentials

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Proving that sin⁡(z+w)=sin⁡(w)cos⁡(z)+sin⁡(z)cos⁡(w)  using complex exponentials

Luciana Cline · Accepted Answer

Consider the RHS,sin⁡zcos⁡w+cos⁡zsin⁡w=eiz−e−iz2ieiw+e−iw2+eiw−e−iw2ieiz+e−iz2[By doing the usual calculation which is skipped]=2ei(z+w)−e−i(z+w)4i=ei(z+w)−e−i(z+w)2i=sin⁡(z+w)Proved.

Mercedes Chang · Accepted Answer

sin⁡(z+w)=exp⁡(i(z+w))−exp⁡(−i(z+w))2i=exp⁡(iz)exp⁡(iw)−exp⁡(−iz)exp⁡(−iw)2i=exp⁡(iz)exp⁡(iw)−exp⁡(iz)exp⁡(−iw)+exp⁡(iz)exp⁡(−iw)−exp⁡(−iz)exp⁡(−iw)4i+exp⁡(iz)exp⁡(iw)−exp⁡(−iz)exp⁡(iw)+exp⁡(−iz)exp⁡(iw)−exp⁡(−iz)exp⁡(−iw)4i=exp⁡(iz)(exp⁡(iw)−exp⁡(−iw))+exp⁡(−iz)(exp⁡(iw)−exp⁡(−iw))4i+exp⁡(−iw)(exp⁡(iz)−exp⁡(−iz))+exp⁡(iw)(exp⁡(iz)−exp⁡(−iz))4i=(exp⁡(iz)+exp⁡(−iz))(exp⁡(iw)−exp⁡(−iw))2⋅2i+(exp⁡(iw)+exp⁡(−iw))(exp⁡(iz)−exp⁡(−iz))2⋅2i=cos⁡(z)sin⁡(w)+cos⁡(w)sin⁡(z)

Proving that \(\displaystyle{\sin{{\left({z}+{w}\right)}}}={\sin{{\left({w}\right)}}}{\cos{{\left({z}\right)}}}+{\sin{{\left({z}\right)}}}{\cos{{\left({w}\right)}}}\) using complex exponentials

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