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

1 Answer
Cesareo R.
Sep 20, 2016

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)#

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