Question

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!

Answer 1

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)`

Answer 2

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 #