Complex numbers with modulus argument

I am stuck on this question for homework:

"Using the fact that if $z = \cos\theta + i \sin\theta$, then $z^n + z^{-n} = 2\cos n\theta$. Show that $$\cos^3 \theta = \tfrac14(\cos3 \theta+3\cos\theta)."$$

Thank you for helping me answering this question.

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First: you can explain why $\frac1{(\cos\,u+i\sin\,u)^3}=\cos\,3u-i\sin\,3u$, right? – Ｊ. M. Nov 8 '12 at 7:22
Welcome to Math.SE. Please add the homework tag to your questions if they are indeed homework. – Harald Hanche-Olsen Nov 8 '12 at 7:25
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Did Nov 8 '12 at 7:44
Hi, thank you for answering my question. To J.M: sorry but i cant explain that answer. To Harald Hanche-Olsen : I will do right now. To did: thank you for the advice. – Zoren Nov 8 '12 at 7:58

Key concept is to apply the angle sum identity to show the desired relationship.

$\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(b)$

$\implies \sin(2x) = \sin(x+x) = \sin(x)\cos(x)+\cos(x)\sin(x)$

$\implies \sin(2x) = 2 \sin(x)\cos(x)$

$\cos(x + y) = \cos(x)\cos(y) - \sin(a)\sin(b)$

$\implies \cos(2x) = \cos(x+x) = \cos(x)^2 - \sin(x)^2$

$\implies \cos(3x) = \cos(1x + 2x) = \cos(2x)\cos(x) - \sin(x)\sin(2x)$

$\implies \cos(3x) = (\cos(x)^2 - \sin(x)^2)\cos(x) - \sin(x)(2 \sin(x)\cos(x))$

$\implies \cos(3x) = \cos(x)^3 - \sin(x)^2\cos(x) - 2\sin^2(x) \cos(x)$

$\implies \cos(3x) = \cos(x)^3 - 3\sin^2(x) \cos(x)$

$\implies \cos(x)^3 = \cos(3x) + 3\sin^2(x) \cos(x)$

$\sin^2(x) + \cos^2(x) = 1$

$\implies \sin^2 = 1 - \cos^2(x)$

$\implies \cos(x)^3 = \cos(3x) + 3 (1 - \cos^2(x)) \cos(x)$

$\implies \cos(x)^3 = \cos(3x) + 3\cos(x) - 3\cos^3(x)$

$\implies 4 \cos(x)^3 = \cos(3x) + 3\cos(x)$

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

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 Hi lewellen, thank you very much for helping with this question. But is there another way in doing it by using the properties z=cosθ+isinθ and z^n+z^−n=2cosnθ? – Zoren Nov 8 '12 at 8:19