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 ?

Thanks!

2 Answers
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)

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