## 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 $4x^2-2x-1=0$

1. anonymous

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

2. anonymous

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

3. Mertsj

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

4. anonymous

Oh right.

5. anonymous

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

so there must be some connection here.

7. anonymous

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

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

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

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

11. anonymous

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

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

13. anonymous

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

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

i meant of course i replaced all the sines by cosines

16. anonymous

oh i see my mistake.

17. anonymous

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

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

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