How do I use z^n=r^n cis(n(θ+2πk)) to find the exact value of cos(2π/5)?

Question

zepdrix · Answer

hmmmmmmmmm

anonymous · Answer

So im just researching about de Moivre's Theorem now..

anonymous · Answer

cis?  Does that mean theta was born an angle and is content with being an angle?

zepdrix · Answer

-_-

zepdrix · Answer

I don't even understand what you just said wio :p

anonymous · Answer

oh cis is just the term we use for (cos theta+isin theta) sorry for the confusion

anonymous · Answer

Oh ok.

anonymous · Answer

I suppose you want to let $r=1$?  Not sure yet.

anonymous · Answer

$n=2/5$?

anonymous · Answer

im looking at http://answers.yahoo.com/question/index?qid=20081214120934AA11Edn and it looks similar to what im having trouble with

zepdrix · Answer

$$\Large\bf\sf \cos\left(\frac{2\pi}{5}\right)\quad=\quad \operatorname{\Re}\left(\cos\frac{2\pi}{5}+\mathcal i \sin\frac{2\pi}{5}\right)$$Do we want to do this maybe?
Use the "real part" notation to get it into complex form.
Then maybe we can apply De Moivre's Theorem or something...$$\Large\bf\sf \operatorname{\Re}\left(\cos 2\pi+\mathcal i \sin 2\pi\right)^{1/5}$$Mmmmmmmmmmmmmmmmmm I dunno...
Just throwing ideas out.

zepdrix · Answer

I could explain the one in the yahoo answers lol.
But I don't quite understand this problem XD Grr

anonymous · Answer

ah, this website has it all me thinks https://suite101.com/a/de-moivres-theorem-examples-cos3x-sin3x-cos4x-and-sin4x-a271768

anonymous · Answer

when you expand z^5
It said to use z=x+yi and then expand (x+yi)^5

anonymous · Answer

Okay, how about this... let $n = 1/5$.