How does #(costheta_1costheta_2-sintheta_1sintheta_2)+i(costheta_1sintheta_2+sintheta_1costheta_2)# become: #[cos(theta_1+theta_2)+isin(theta_1+theta_2)]#?

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Answer 1 · 2017-07-26T04:51:57

Please see below.

Recall #cos(A+B)=cosAcosB-sinAsinB# and #sin(A+B)=sinAcosB+cosAsinB#

Hence #cos(theta_1+theta_2)=costheta_1costheta_2-sintheta_1sintheta_2#

and #sin(theta_1+theta_2)=sintheta_1costheta_2+costheta_1sintheta_2#

= #costheta_1sintheta_2+sintheta_1costheta_2#

Hence #(costheta_1costheta_2-sintheta_1sintheta_2)-i(costheta_1sintheta_2+sintheta_1costheta_2)#

becomes #[cos(theta_1+theta_2)+isin(theta_1+theta_2)#

Answer 2 · 2017-07-26T07:53:22

See below.

Using de Moivre's identity

#e^(i theta) = cos theta+i sin theta# we have

#e^(i theta_1) e^(i theta_2) = (cos theta_1+i sin theta_1)(cos theta_2+i sin theta_2) = e^(i(theta_1+theta_2)) =#

#=cos(theta_1+theta_2)+i sin(theta_1+theta_2)#