Question

How do you use Demoivre's theorem to show that `sin6theta=sin2theta(3-4sin^2theta)(1-4sin^2theta)`?

Answer 1

See below.

Explanation:

de Moivre's identity states that

`e^(ix)=cos(ix)+isin(ix)` so taking

`t_1=1-4sin^2theta =(1+2sintheta)(1-2sintheta) = (1-(e^(itheta)-e^(-itheta))/i) (1+(e^(itheta)-e^(-itheta))/i)`
`t_2=3-4sin^2theta=(sqrt(3)+2sintheta)(sqrt(3)-2sintheta)= (sqrt(3)-(e^(itheta)-e^(-itheta))/i) (sqrt(3)+(e^(itheta)-e^(-itheta))/i)`
and
`t_3=(e^(i2theta)-e^(-i2theta))/(2i)`

we have

`t_1=e^(2itheta)+e^(-2itheta)-1`
`t_2=e^(2itheta)+e^(-2itheta)+1`

and after multilying `t_1t_2t_3`

`t_1t_2t_3=(e^(6itheta)-e^(-6itheta))/(2i) = sin(6theta)`