How does $({cos θ}_{1} {cos θ}_{2} - {sin θ}_{1} {sin θ}_{2}) + i ({cos θ}_{1} {sin θ}_{2} + {sin θ}_{1} {cos θ}_{2})$ $(costheta_1costheta_2-sintheta_1sintheta_2)+i(costheta_1sintheta_2+sintheta_1costheta_2)$ become: $[cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2})]$ $[cos(theta_1+theta_2)+isin(theta_1+theta_2)]$ ?

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How does $({cos θ}_{1} {cos θ}_{2} - {sin θ}_{1} {sin θ}_{2}) + i ({cos θ}_{1} {sin θ}_{2} + {sin θ}_{1} {cos θ}_{2})$ $(costheta_1costheta_2-sintheta_1sintheta_2)+i(costheta_1sintheta_2+sintheta_1costheta_2)$ become: $[cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2})]$ $[cos(theta_1+theta_2)+isin(theta_1+theta_2)]$ ?

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Answer 1 · 2017-07-26T04:51:57

Please see below.

Explanation:

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

Hence $cos (θ_{1} + θ_{2}) = {cos θ}_{1} {cos θ}_{2} - {sin θ}_{1} {sin θ}_{2}$ $cos(theta_1+theta_2)=costheta_1costheta_2-sintheta_1sintheta_2$

and $sin (θ_{1} + θ_{2}) = {sin θ}_{1} {cos θ}_{2} + {cos θ}_{1} {sin θ}_{2}$ $sin(theta_1+theta_2)=sintheta_1costheta_2+costheta_1sintheta_2$

= ${cos θ}_{1} {sin θ}_{2} + {sin θ}_{1} {cos θ}_{2}$ $costheta_1sintheta_2+sintheta_1costheta_2$

Hence $({cos θ}_{1} {cos θ}_{2} - {sin θ}_{1} {sin θ}_{2}) - i ({cos θ}_{1} {sin θ}_{2} + {sin θ}_{1} {cos θ}_{2})$ $(costheta_1costheta_2-sintheta_1sintheta_2)-i(costheta_1sintheta_2+sintheta_1costheta_2)$

becomes $[cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2})$ $[cos(theta_1+theta_2)+isin(theta_1+theta_2)$

Answer 2 · 2017-07-26T07:53:22

See below.

Explanation:

Using de Moivre's identity

$e^{i θ} = cos θ + i sin θ$ $e^(i theta) = cos theta+i sin theta$ we have

$e_{1}^{i θ_{1}} e_{2}^{i θ_{2}} = ({cos θ}_{1} + i {sin θ}_{1}) ({cos θ}_{2} + i {sin θ}_{2}) = e^{i (θ_{1} + θ_{2})} =$ $e^(i theta_1) e^(i theta_2) = (cos theta_1+i sin theta_1)(cos theta_2+i sin theta_2) = e^(i(theta_1+theta_2)) =$

$= cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2})$ $=cos(theta_1+theta_2)+i sin(theta_1+theta_2)$

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