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satellite73
 4 years ago
ok let me ask the question this way.
why (besides the fact that it is true) is
\[\cos\left(\frac{\pi}{5}\right)\] the same as the solution to
\[4x^22x1=0\]
satellite73
 4 years ago
ok let me ask the question this way. why (besides the fact that it is true) is \[\cos\left(\frac{\pi}{5}\right)\] the same as the solution to \[4x^22x1=0\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe this is just not that interesting, but it is bugging me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't get what you mean. \(\cos(\frac{\pi}{5})\) is not a root for \(4x^2−2x−1\).

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Oh but it is. One does not lightly challenge the Great Satellite.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol, i only know it is because i solved \[4x^22x1=0\], got \[x=\frac{1\pm\sqrt{5}}{4}\] and then took the inverse cosine and got \[\frac{\pi}{5},\frac{3\pi}{5}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so there must be some connection here.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In other words, you're asking how we can show that either \(\large \frac{1\pm \sqrt{5}}{4}=\cos(\frac{\pi}{5})\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i know it is true, i am just wondering what the connection between the quadratic and \[\frac{\pi}{5}\] is. maybe something to do with tenth roots of 1, maybe

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Write \( z = e^{i\pi/5} = \cos(\pi/5) + i\sin(\pi/5) \) and \( c = \cos(\pi/5), s = \sin(\pi/5) \). Then \[1 = z^5 = ( c + is)^5 \] Now expand that out, collect all the imaginary and/or real terms. I'm sure after some manipulationincluding use of \( c^2 + s^2 = 1 \)you'll be able to show that \[ 4c^2  2c  1 = 0 . \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2**correction, 1 = z^5 = ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i thought tenth roots so now i am happier. but let us suppose we were asked to find \[\cos(\frac{\pi}{5})\] is there something we would use to end up with the quadratic?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think this has something to do with what we're discussing. http://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0could be i just tried the method above and got a huge mess. equated the real part of \[(\cos(\frac{\pi}{5})+i\sin(\frac{\pi}{5}))^5\]to 1, replaced all the cosines by sines and got a huge mess (degree 10)

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Shouldn't be, for example: http://www.blog.republicofmath.com/archives/3738 Notice at the end you'll have in its most obvious form, \[ cos(\pi/5) = \sqrt{\frac{3 + \sqrt{5}}{8}} \] but this is the same as your earlier expression \( (1 + \sqrt{5})/4 \)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i meant of course i replaced all the sines by cosines

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is also a typo in what you sent, but no matter, i forgot i get 1 on the left as well as on the right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and now i am wondering why, in the link you sent, since they are trying to find cosine, they solved for sine first. why not just write in term of cosine to begin with?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and now i see why, he gets to divide by sine because it is set equal zero, not 1.
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