How do you use Demoivre's theorem to show that sin6θ=sin2θ(34sin2θ)(14sin2θ)?

1 Answer
Sep 20, 2016

See below.

Explanation:

de Moivre's identity states that

eix=cos(ix)+isin(ix) so taking

t1=14sin2θ=(1+2sinθ)(12sinθ)=(1eiθeiθi)(1+eiθeiθi)
t2=34sin2θ=(3+2sinθ)(32sinθ)=(3eiθeiθi)(3+eiθeiθi)
and
t3=ei2θei2θ2i

we have

t1=e2iθ+e2iθ1
t2=e2iθ+e2iθ+1

and after multilying t1t2t3

t1t2t3=e6iθe6iθ2i=sin(6θ)