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

Sep 20, 2016

See below.

#### Explanation:

de Moivre's identity states that

${e}^{i x} = \cos \left(i x\right) + i \sin \left(i x\right)$ so taking

${t}_{1} = 1 - 4 {\sin}^{2} \theta = \left(1 + 2 \sin \theta\right) \left(1 - 2 \sin \theta\right) = \left(1 - \frac{{e}^{i \theta} - {e}^{- i \theta}}{i}\right) \left(1 + \frac{{e}^{i \theta} - {e}^{- i \theta}}{i}\right)$
${t}_{2} = 3 - 4 {\sin}^{2} \theta = \left(\sqrt{3} + 2 \sin \theta\right) \left(\sqrt{3} - 2 \sin \theta\right) = \left(\sqrt{3} - \frac{{e}^{i \theta} - {e}^{- i \theta}}{i}\right) \left(\sqrt{3} + \frac{{e}^{i \theta} - {e}^{- i \theta}}{i}\right)$
and
${t}_{3} = \frac{{e}^{i 2 \theta} - {e}^{- i 2 \theta}}{2 i}$

we have

${t}_{1} = {e}^{2 i \theta} + {e}^{- 2 i \theta} - 1$
${t}_{2} = {e}^{2 i \theta} + {e}^{- 2 i \theta} + 1$

and after multilying ${t}_{1} {t}_{2} {t}_{3}$

${t}_{1} {t}_{2} {t}_{3} = \frac{{e}^{6 i \theta} - {e}^{- 6 i \theta}}{2 i} = \sin \left(6 \theta\right)$