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Find 5th roots of unity solving x^5 -1=0 and use the result for sin18 and cos18

  1. Jun 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Find 5th roots of unity solving algebraically x^5-1=0. Using the result, find sin18 and cos18


    3. The attempt at a solution

    [itex]
    x^5 = 1\\
    x = \sqrt[5]{1}[/itex]

    since we have 5 roots:
    [itex]
    x_k, k = 0,1,2,3,4 \\ \\

    x_k = e^{i\frac{2k\pi}{n}}, n=5 \\ x_0 = e^{i0} = 1\\ x_1 = e^{i\frac{2\pi}{5}} = cos\frac{2\pi}{5} + isin\frac{2\pi}{5} \\ x_2 = e^{i\frac{4\pi}{5}} = cos\frac{4\pi}{5} + isin\frac{4\pi}{5} \\ x_3 = e^{i\frac{6\pi}{5}} = cos\frac{6\pi}{5} + isin\frac{6\pi}{5} \\ x_5 = e^{i\frac{8\pi}{5}} = cos\frac{8\pi}{5} + isin\frac{8\pi}{5}[/itex]

    now how do I find sin18 and cos18??
     
  2. jcsd
  3. Jun 20, 2012 #2
    Perhaps you could use de Moivre's formula
     
  4. Jun 20, 2012 #3
    Hi tonit! :smile:

    You have [itex] x_1 = e^{72i} = cos72 + isin72[/itex], where the angle is in degrees.

    Can you express these in terms of cosine and sine 18? Then use the binomial expansion to the index 5 of the root you get(in terms of cos and sin 18) and equate the imaginary and real coefficients. You will get two equations, one of which you can easily solve for their values.
     
  5. Jun 20, 2012 #4
    [itex] e^{i\frac{2\pi }{5}} = sin(18) + icos18[/itex] right?
     
  6. Jun 20, 2012 #5
    Yes!
     
  7. Jun 20, 2012 #6
    I got

    [itex]sin^5θ - 10sin^3θ cos^2θ + 5sinθ cos^4θ = 1[/itex]
    and
    [itex]i5sin^4θcosθ - i10sin^2θcos^3θ + icos^5θ = 0[/itex]

    where [itex] θ = \frac{\pi}{10} = 18^{\circ}[/itex]

    is this ok?
     
    Last edited: Jun 20, 2012
  8. Jun 20, 2012 #7
    Yes! :smile:

    Now try solving one of these equations, as a single trigonometric ratio. Which one would be easier for you to solve? :wink:
     
  9. Jun 20, 2012 #8
    I guess the first one would be easier,

    and I got

    [itex]16sin^5\theta - 20sin^3\theta + 5sin\theta = 1[/itex]

    I'm stuck again.....:@
     
  10. Jun 20, 2012 #9
    Actually...the second one would be easier. You have a zero in the RHS, so you can divide the equation by icosθ and get rid of a worry :wink:
     
  11. Jun 20, 2012 #10
    alright so after simplifying I get [itex]16sin^4\theta - 12sin^2\theta + 1 = 0[/itex]

    now it's pretty obvious to solve. thanks :D
     
  12. Jun 20, 2012 #11
    I would use the multiple-angle identity.
     
  13. Jun 20, 2012 #12
    Hi dimension10!
    Could you please explain how, keeping in mind the OP was asked to derive the result using the roots of unity?
     
  14. Jun 20, 2012 #13
    Oh, I was thinking that he wanted help on finding the 5th roots of unity using the cosine of pi/5 and sine of pi/5.
     
  15. Jun 20, 2012 #14
    Yes, but just remember you will need to reject one of the values you get as,

    [tex]0 \leq sin^2\theta \leq 1[/tex]
     
  16. Jun 20, 2012 #15
    yeah, I'll keep that in my mind :smile:
     
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