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Prove $x^2+\frac{1}{x^2}=2\cos(2\theta)$ and $x^3+\frac{1}{x^3}=2\cos(3\theta)$ knowing that there exist a number $x$ given angle $\theta$ such that $x+\frac{1}{x}=2\cos(\theta)$

Doesn't really know how to start this problem, thought that I would some how need to use the double angle identities

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  • $\begingroup$ The number $x$ is necessarily real? $\endgroup$ – José Carlos Santos Dec 12 '19 at 9:26
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    $\begingroup$ Have you tried squaring both sides? $\endgroup$ – T.J. Gaffney Dec 12 '19 at 9:28
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$$\left(x+\frac 1x\right)^2 = x^2+2+\frac 1{x^2} = 4\cos^2\theta \\\implies x^2+\frac{1}{x^2} = 2(2\cos^2\theta-1) = 2\cos2\theta$$

Similarly

$$\left(x+\frac 1x\right)^3 = x^3 + 3\left(x+\frac 1x\right) + \frac{1}{x^3} = x^3+\frac{1}{x^3} + 6\cos\theta = 8\cos^3\theta$$

$$ \implies x^3+\frac{1}{x^3} = 2(4\cos^3\theta-3\cos\theta) = 2\cos3\theta$$

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    $\begingroup$ My sincere congratulations. $\endgroup$ – Sebastiano Dec 12 '19 at 9:41
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    $\begingroup$ @Sebastiano Thank you! $\endgroup$ – Ak. Dec 12 '19 at 9:42
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    $\begingroup$ +1 for a beautiful answer! $\endgroup$ – Toby Mak Dec 12 '19 at 9:45
  • $\begingroup$ Thank you!! @TobyMak $\endgroup$ – Ak. Dec 12 '19 at 9:53
  • $\begingroup$ Nice and elegant (+1) $\endgroup$ – trancelocation Dec 12 '19 at 11:57
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$$x=\cos\theta\pm i\sin\theta $$

Taking $+$ sign for integer $n$,

using https://en.m.wikipedia.org/wiki/De_Moivre's_formula,

$$x^n=\cos n\theta+i\sin n\theta$$

$$1/x^n=?$$

Similarly consider $-$ sign

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