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

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

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Answer 1 · 2016-09-20T08:44:42

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

1 Answer

Explanation:

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Science

Math

Humanities

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

1 Answer

Explanation:

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