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

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

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

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