How do you use Demoivre's theorem to show that sin6theta=sin2theta(3-4sin^2theta)(1-4sin^2theta)sin6θ=sin2θ(34sin2θ)(14sin2θ)?

1 Answer
Sep 20, 2016

See below.

Explanation:

de Moivre's identity states that

e^(ix)=cos(ix)+isin(ix)eix=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)t1=14sin2θ=(1+2sinθ)(12sinθ)=(1eiθeiθi)(1+eiθeiθi)
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)t2=34sin2θ=(3+2sinθ)(32sinθ)=(3eiθeiθi)(3+eiθeiθi)
and
t_3=(e^(i2theta)-e^(-i2theta))/(2i)t3=ei2θei2θ2i

we have

t_1=e^(2itheta)+e^(-2itheta)-1t1=e2iθ+e2iθ1
t_2=e^(2itheta)+e^(-2itheta)+1t2=e2iθ+e2iθ+1

and after multilying t_1t_2t_3t1t2t3

t_1t_2t_3=(e^(6itheta)-e^(-6itheta))/(2i) = sin(6theta)t1t2t3=e6iθe6iθ2i=sin(6θ)