Express $\cos(\pi/5)$ in terms of a sum of powers of the principal 100th root of unity.

Question

Express $\cos(\pi/5)$ in terms of a sum of powers of the principal $100{th}$ root of unity.

Using the formula, $w_n = \cos(2\pi/n) + i \sin(2\pi/n)$

I have calculated, $w_{100} = \cos(\pi/50) + i \sin(\pi/50)$

$w_{100}^{10} = \cos(\pi/5) + i \sin(\pi/5)$

But $\cos(\pi/5)^{10} \neq w_{100}^{10}$, so I'm confused how to progress this problem.

Regards.

Note the sums of powers in the question. What's $w_n+w_n^{-1}$ — Gerry Myerson, Mar 19 '14 at 12:52

JP McCarthy · Accepted Answer · 2014-03-19T13:10:30.073

1

$$\frac12\left(w_{100}^{10}+w_{100}^{90}\right).$$

This works because

$$\begin{align} w_{100}&=\cos(\pi/50)+i\sin(\pi/50) \\ \Rightarrow w_{100}^{10}&=\cos(\pi/5)+i\sin(\pi/5) \\ \Rightarrow w_{100}^{90}&=\cos(9\pi /5)+i\sin(9\pi/5) \\ &=\cos(-\pi/5)+i\sin(-\pi/5) \\ \Rightarrow \frac{1}{2}\left(w_{100}^{10}+w_{100}^{90}\right)&=\frac{1}{2}\left(\cos(\pi/5)+i\sin(\pi/5)+\cos(-\pi/5)+i\sin(-\pi/5)\right) \end{align}$$

Now what do you know about $\cos(-A)$ and $\sin(-A)$?

edited Mar 19 '14 at 13:10

answered Mar 19 '14 at 12:52

JP McCarthy

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I dont think this is correct. I'm still left with imaginary parts in the solution from the $i \sin()$ bit. – Danny Rancher Mar 19 '14 at 13:05
No you don't... look at the edit. – JP McCarthy Mar 19 '14 at 13:10
OK. I understand completely now. Thank you. I had forgotten that trig identity. Is it possible to express in terms of $\sin(\pi/5)$ as well? I've experimented with different coefficients and the same powers but I cannot remove the imaginary element. – Danny Rancher Mar 19 '14 at 15:49
$\Im(z)=\frac{1}{2i}(z-\bar{z})$ – JP McCarthy Mar 19 '14 at 16:10

Express $\cos(\pi/5)$ in terms of a sum of powers of the principal 100th root of unity.

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