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I'm struggling with a question which requires the use of the De Moivre identity. The question is as follows.

Prove that $$8(\cos{^6}\theta + \sin{^6}\theta) = 3\cos 4\theta + 5$$

[Hint: write $\cos\theta$ and $\sin\theta$ in terms of $e^{i\theta}$ and $e^{-i\theta}$]

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I've gotten as far as writing out $\cos{^6}\theta$ as $[\frac{1}{2}(e^{i\theta}+e^{-i\theta}]^6$ and $\sin{^6}\theta$ as $[\frac{1}{2i}(e^{i\theta}-e^{-i\theta}]^6$. From here I'm not sure where to go, as I'm not sure how to use De Moivre's theorem on the right hand side to write $\cos 4\theta$ in terms of complex numbers. Any hint on where to go from here would be greatly appreciated.

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  • $\begingroup$ Simply carefully then you will get $\frac{3e^{-4iθ}+3e^{4iθ}+10}{2}$ which is your desire answer $\endgroup$
    – emonHR
    Dec 25 '19 at 19:19
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$\left(\frac{e^{i\theta \:}+e^{-i\theta \:}}{2}\right)^6+\left(\frac{e^{i\theta \:\:}-e^{-i\theta \:\:}}{2i}\right)^6=\frac{\left(e^{iθ}+e^{-iθ}\right)^6}{64}+\frac{\left(e^{iθ}-e^{-iθ}\right)^6}{-64}=\frac{\left(e^{iθ}+e^{-iθ}\right)^6-\left(e^{iθ}-e^{-iθ}\right)^6}{64}$
Apply binomial theorem, $$\left(e^{iθ}+e^{-iθ}\right)^6=e^{6iθ}+6e^{4iθ}+15e^{2iθ}+20+15e^{-2iθ}+6e^{-4iθ}+e^{-6iθ}$$ $$\left(e^{iθ}-e^{-iθ}\right)^6=\left(e^{6iθ}-6e^{4iθ}+15e^{2iθ}-20+15e^{-2iθ}-6e^{-4iθ}+e^{-6iθ}\right)$$ \begin{align}8\left(\frac{\left(e^{iθ}+e^{-iθ}\right)^6-\left(e^{iθ}-e^{-iθ}\right)^6}{64}\right)&=8\left(\frac{12e^{-4iθ}+12e^{4iθ}+40}{64}\right)\\ &=8\left(\frac{3e^{-4iθ}+3e^{4iθ}+10}{16}\right)\\ &=\frac{3e^{-4iθ}+3e^{4iθ}+10}{2}\\ &=3.\frac{e^{-4iθ}+e^{4iθ}}{2}+5\\ &=3\cos4θ+5 \end{align}

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  • $\begingroup$ Really appreciate seeing this approach, thanks a lot $\endgroup$
    – Ataa Altaf
    Dec 25 '19 at 20:05
  • $\begingroup$ I'm glad I could help $\endgroup$
    – emonHR
    Dec 26 '19 at 6:32
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There are easier ways to solve this, for example $$\cos^6(\theta)+\sin^6(\theta) = \left( \cos^2(\theta)+\sin^2(\theta) \right)^3 - 3 \cos^2(\theta)\sin^2(\theta) \left( \cos^2(\theta)+\sin^2(\theta) \right) = 1 - 3\cos^2(\theta)\sin^2(\theta) = 1-\frac{3}{4}\sin^2 (2\theta) = 1-\frac{3}{4}\left( \frac{1-\cos(4\theta)}{2} \right) = \frac{5+3\cos(4\theta)}{8}$$

but if we insist on Abe. De Moivre's theorem, we can go like this

$$\left[\frac{1}{2} \left( e^{i\theta}+e^{-i\theta} \right) \right]^6 + \left[\frac{1}{2i} \left( e^{i\theta}-e^{-i\theta} \right) \right]^6 = \frac{1}{64} \left[ \left( e^{i\theta}+e^{-i\theta} \right)^6 - \left( e^{i\theta}-e^{-i\theta} \right)^6 \right] = \frac{1}{64} \left( \left( e^{i\theta}+e^{-i\theta} \right)^3 - \left( e^{i\theta}-e^{-i\theta} \right)^3 \right)\left( \left( e^{i\theta}+e^{-i\theta} \right)^3 + \left( e^{i\theta}-e^{-i\theta} \right)^3 \right) = \frac{1}{64} \left( 6e^{i\theta} + 2e^{-3i\theta} \right)\left( 6e^{-i\theta} + 2e^{3i\theta} \right) = \frac{1}{64} \left( 36 + 12e^{4i\theta} + 12e^{-4i\theta} +4 \right) = \frac{1}{64} \left( 40 + 24\times\frac{1}{2} \left( e^{4i\theta}+e^{-4i\theta} \right) \right) = \frac{1}{64}\left( 40 + 24\cos(4\theta)\right) = \frac{5+3\cos(4\theta)}{8}$$

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  • $\begingroup$ I think it should be $\frac{3e^{-4iθ}+3e^{4iθ}+10}{16}$. Please recheck your answer $\endgroup$
    – emonHR
    Dec 25 '19 at 19:22
  • $\begingroup$ Rechecked it, pretty sure the answer is correct, thanks for heads up by the way, it's always better to double check. $\endgroup$ Dec 25 '19 at 19:26
  • $\begingroup$ Shouldn't it be $\color{red}+3\cos 4\theta$? $\endgroup$ Dec 25 '19 at 19:38
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    $\begingroup$ Awww, minus sign must be plus, rushing at last step, thanks man! $\endgroup$ Dec 25 '19 at 19:44
  • $\begingroup$ Thanks very much all of you, really helpful! Merry christmas $\endgroup$
    – Ataa Altaf
    Dec 25 '19 at 20:03
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To make the notations (and the computations) lighter, set $u=\mathrm e^{i\theta}$, so that $\bar u=\mathrm e^{-i\theta}$, $u\,\bar u=1$ and $u^k+\bar u^k=2\cos k\theta$. We have, using high-school identities, \begin{align} 8(\cos^6 \theta + \sin^6\theta)& =8\biggl(\frac{(u+\bar u)^6}{64}+ \frac{(u-\bar u)^6}{-64}\biggr)=\frac18\Bigl((u+\bar u)^6-(u-\bar u)^6\Bigr)\\ &=\frac18\Bigl[\Bigl(u^6+6u^4+15u^2+20+15\bar u^2+6\bar u^4+\bar u^6\Bigr)\\ &\phantom{=}\qquad-\Bigl(u^6-6u^4+15u^2-20+15\bar u^2-6\bar u^4+\bar u^6\Bigr)\Bigr] \\ &= \frac18\Bigl[12(u^4+\bar u^4)+40\Bigr]=\frac18\Bigl[24\cos 4\theta +40\Bigr]=3\cos4\theta+5. \end{align}

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  • $\begingroup$ Your notation makes the problem so much less of an eyesore, many thanks and merry Christmas! $\endgroup$
    – Ataa Altaf
    Dec 25 '19 at 20:06

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