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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 \[4x^2-2x-1=0\]

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  1. anonymous
    • 4 years ago
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    maybe this is just not that interesting, but it is bugging me

  2. anonymous
    • 4 years ago
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    I don't get what you mean. \(\cos(\frac{\pi}{5})\) is not a root for \(4x^2−2x−1\).

  3. Mertsj
    • 4 years ago
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    Oh but it is. One does not lightly challenge the Great Satellite.

  4. anonymous
    • 4 years ago
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    Oh right.

  5. anonymous
    • 4 years ago
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    lol, i only know it is because i solved \[4x^2-2x-1=0\], got \[x=\frac{1\pm\sqrt{5}}{4}\] and then took the inverse cosine and got \[\frac{\pi}{5},\frac{3\pi}{5}\]

  6. anonymous
    • 4 years ago
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    so there must be some connection here.

  7. anonymous
    • 4 years ago
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    In other words, you're asking how we can show that either \(\large \frac{1\pm \sqrt{5}}{4}=\cos(\frac{\pi}{5})\).

  8. anonymous
    • 4 years ago
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    no 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

  9. JamesJ
    • 4 years ago
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    Write \( z = e^{i\pi/5} = \cos(\pi/5) + i\sin(\pi/5) \) 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 manipulation--including use of \( c^2 + s^2 = 1 \)--you'll be able to show that \[ 4c^2 - 2c - 1 = 0 . \]

  10. JamesJ
    • 4 years ago
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    **correction, -1 = z^5 = ....

  11. anonymous
    • 4 years ago
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    i 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?

  12. anonymous
    • 4 years ago
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    I think this has something to do with what we're discussing. http://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)

  13. anonymous
    • 4 years ago
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    could 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)

  14. JamesJ
    • 4 years ago
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    Shouldn'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 \( (1 + \sqrt{5})/4 \)

  15. anonymous
    • 4 years ago
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    i meant of course i replaced all the sines by cosines

  16. anonymous
    • 4 years ago
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    oh i see my mistake.

  17. anonymous
    • 4 years ago
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    there 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.

  18. anonymous
    • 4 years ago
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    and 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?

  19. anonymous
    • 4 years ago
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    and now i see why, he gets to divide by sine because it is set equal zero, not -1.

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