Prove #((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx#?

Prove ((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx ?

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2 Answers
Cem Sentin
Dec 17, 2017

Explanation is below

Explanation:

#(1+cos2x+isin2x)/(1+cos2x-isin2x)#

=#[2(cosx)^2+2i*sinx*cosx]/[2(cosx)^2-2i*sinx*cosx]#

=#[2cosx*(cosx+isinx)]/[2cosx*(cosx-isinx)]#

=#(cosx+isinx)/(cosx-isinx)#

=#(cosx+isinx)^2/[(cosx-isinx)*(cosx+i*sinx)]#

=#[(cosx)^2-(sinx)^2+2i*sinx*cosx]/[(cosx)^2+(sinx)^2]#

=#(cos2x+isin2x)/1#

=#cos2x+isin2x#

Thus,

#[(1+cos2x+isin2x)/(1+cos2x-isin2x)]^n#

=#(cos2x+isin2x)^n#

=#cos(2nx)+isin(2nx)#

Cesareo R.
Dec 17, 2017

See below.

Explanation:

#1+e^(i2x) = e^(ix)(e^(ix)+e^(-ix))#
#1+e^(-i2x) = e^(-ix)(e^(ix)+e^(-ix))# so

#((1+e^(i2x))/(1+e^(-i2x)))^n = (e^(i2x))^n = e^(i2nx) = cos(2nx)+isin(2nx)#

NOTE

We were using the de Moivre's identity

#e^(i phi) = cos phi + i sin phi #

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