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I have trouble doing this exercise:

Let $z = cosθ + isinθ$ Expand $(z + z^{−1})^6 and (z-z^{−1})^6$ Hence show that $cos^6θ+sin^6θ=1/8(3cos4θ+5)$

I expanded these two brackets and I got: $2cos6θ+12cos4θ+30cos2θ+20$ for the first one and $2cos6θ-12cos4θ+30cos2θ-20$ for the second one

Is that correct?

I also know that $(z + z^{−1})^6$ is $64cos^6θ$ and $(z-z^{−1})^6$ is $64sin^6θ$

but when I add these two result it doesn't work

Where do I make the mistake?

Thanks for help!

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    $\begingroup$ I reckon that $(z-z^{-1})^6$ is actually $-64\sin^6 \theta$, $\endgroup$ May 15 '17 at 17:31
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$$z-z^{-1}=2i\sin\theta\implies(z-z^{-1})^6=(2\sin\theta)^6i^6=-64\sin^6\theta$$

$$64(\cos^6\theta+\sin^6\theta)=(z+z^{-1})^6-(z-z^{-1})^6=2\left[\binom61\left(z^4+z^{-4}\right)+\binom63\right]$$

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  • $\begingroup$ Use en.wikipedia.org/wiki/De_Moivre%27s_formula $\endgroup$ May 15 '17 at 17:38
  • $\begingroup$ I don't really understand the second line of your answer. So far I got: 24cos4θ+40 which gives us 8(3cos4θ+5) and not 1/8(3cos4θ+5). Could tell me what I'm doing wrong? ;) $\endgroup$
    – Markowska
    May 15 '17 at 17:47
  • $\begingroup$ @Markowska, Have you noticed $64$ in the left hand side? $\endgroup$ May 15 '17 at 17:48
  • $\begingroup$ Oh! yes, now i can divide it by 64. Thanks a lot $\endgroup$
    – Markowska
    May 15 '17 at 17:49

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