Question

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

Answer 1

maybe this is just not that interesting, but it is bugging me

Answer 2

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

Answer 3

Oh but it is. One does not lightly challenge the Great Satellite.

Answer 4

Oh right.

Answer 5

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
\[\frac{\pi}{5},\frac{3\pi}{5}\]

Answer 6

so there must be some connection here.

Answer 7

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

Answer 8

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

Answer 9

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
\[ 4c^2 - 2c - 1 = 0 . \]

Answer 10

**correction, -1 = z^5 = ....

Answer 11

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?

Answer 12

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

Answer 13

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)

Answer 14

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 \( (1 + \sqrt{5})/4 \)

Answer 15

i meant of course i replaced all the sines by cosines

Answer 16

oh i see my mistake.

Answer 17

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.

Answer 18

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?

Answer 19

and now i see why, he gets to divide by sine because it is set equal zero, not -1.