How do you use Demoivre's theorem to show that $sin 6 θ = sin 2 θ (3 - 4 {sin}^{2} θ) (1 - 4 {sin}^{2} θ)$ $sin6theta=sin2theta(3-4sin^2theta)(1-4sin^2theta)$ ?

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How do you use Demoivre's theorem to show that $sin 6 θ = sin 2 θ (3 - 4 {sin}^{2} θ) (1 - 4 {sin}^{2} θ)$ $sin6theta=sin2theta(3-4sin^2theta)(1-4sin^2theta)$ ?

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Answer 1 · 2016-09-20T08:44:42

See below.

Explanation:

de Moivre's identity states that

$e^{i x} = cos (i x) + i sin (i x)$ $e^(ix)=cos(ix)+isin(ix)$ so taking

$t_{1} = 1 - 4 {sin}^{2} θ = (1 + 2 sin θ) (1 - 2 sin θ) = (1 - \frac{e^{i θ} - e^{- i θ}}{i}) (1 + \frac{e^{i θ} - e^{- i θ}}{i})$ $t_1=1-4sin^2theta =(1+2sintheta)(1-2sintheta) = (1-(e^(itheta)-e^(-itheta))/i) (1+(e^(itheta)-e^(-itheta))/i)$
$t_{2} = 3 - 4 {sin}^{2} θ = (\sqrt{3} + 2 sin θ) (\sqrt{3} - 2 sin θ) = (\sqrt{3} - \frac{e^{i θ} - e^{- i θ}}{i}) (\sqrt{3} + \frac{e^{i θ} - e^{- i θ}}{i})$ $t_2=3-4sin^2theta=(sqrt(3)+2sintheta)(sqrt(3)-2sintheta)= (sqrt(3)-(e^(itheta)-e^(-itheta))/i) (sqrt(3)+(e^(itheta)-e^(-itheta))/i)$
and
$t_{3} = \frac{e^{i 2 θ} - e^{- i 2 θ}}{2 i}$ $t_3=(e^(i2theta)-e^(-i2theta))/(2i)$

we have

$t_{1} = e^{2 i θ} + e^{- 2 i θ} - 1$ $t_1=e^(2itheta)+e^(-2itheta)-1$
$t_{2} = e^{2 i θ} + e^{- 2 i θ} + 1$ $t_2=e^(2itheta)+e^(-2itheta)+1$

and after multilying $t_{1} t_{2} t_{3}$ $t_1t_2t_3$

$t_{1} t_{2} t_{3} = \frac{e^{6 i θ} - e^{- 6 i θ}}{2 i} = sin (6 θ)$ $t_1t_2t_3=(e^(6itheta)-e^(-6itheta))/(2i) = sin(6theta)$

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How do you use Demoivre's theorem to show that $sin 6 θ = sin 2 θ (3 - 4 {sin}^{2} θ) (1 - 4 {sin}^{2} θ)$ $sin6theta=sin2theta(3-4sin^2theta)(1-4sin^2theta)$ ?

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Explanation:

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

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How do you use Demoivre's theorem to show that $sin 6 θ = sin 2 θ (3 - 4 {sin}^{2} θ) (1 - 4 {sin}^{2} θ)$ $sin6theta=sin2theta(3-4sin^2theta)(1-4sin^2theta)$ ?