# 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
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