Evaluate sin(2pi/5)

  • Thread starter Driessen12
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  • #1
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Homework Statement


Show that sin(2pi/5) = ((sqrt(10) + (2sqrt(5)))/4)


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The Attempt at a Solution


i showed that cos(2pi/5) + isin(2pi/5) was a primitive 5th root of unity, and that cos(2pi/5) = (-1 + sqrt(5))/4 but i cannot figure out how to do this one any ideas?
 

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  • #2
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i also can use the fact that x^5 - 1 = 0 and i did a substitution to prove that cos(2pi/5) = (-1 + sqrt(5))/4 using the fact that x^5 - 1 + ( x - 1)(x^4 + x^3 + x^2 + x + 1) and x-1 is nonzero so the second equation on rhs has to be zero and then i used a variable substitution but i don't know how to go about proving the sin(2pi/5) part
 
  • #3
dextercioby
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Well

[tex] \sin^2 \frac{2\pi}{5} + \cos^2 \frac{2\pi}{5} =1 [/tex]

, obviously.
 
  • #5
dextercioby
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How did you get cos(2pi/5) ? I couldn't factor any polynomial.
 
  • #6
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you divide x^4 + x^3 + x^2 + x + 1 by x^2 because it is non zero this is valid. so we get x^2 + x + 1 + 1/x + 1/x^2. then say y = x + 1/x and plugging this back in you get y^2 + y -1 = 0
then use the quadratic formula to get (-1 + sqrt(5))/2. back to y we know that x = cos(2pi/5) + isin(2pi/5) so add that to x^-1 and you end up getting 2cos(2pi/5) = (-1 + sqrt(5))/2
 
  • #8
uart
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Hi bigubau I used a much more mundane method.

Denoting [itex]\frac{2 \pi}{5} = \theta[/itex] we have,

[tex] Re\{(\cos \theta + i \sin \theta)^5\} = 1 [/tex]

[tex] Im\{(\cos \theta + i \sin \theta)^5\} = 0 [/tex]

Expanding the second of these two equations and denoting [itex]x = \cos \theta[/itex] (and using [itex]\sin^2 \theta = 1 - x^2 [/itex] where appropriate) gives,

[tex] i \sin \theta \, \left[ 5x^4 - 10 x^2 (1-x^2) + (1-x^2)^2 \right] = 0[/tex]

which reduces to a quadratic in [itex]x^2[/itex].

[tex] 16x^4 - 12 x^2 + 1 = 0[/tex]

Solving gives,

[tex]x^2 = \frac{6 \pm 2 \sqrt{5}} {16} = \frac{(\sqrt{5} \pm 1)^2}{4^2}[/tex]

[tex]x = \frac{\sqrt{5} - 1}{4}[/tex]

BTW. Something a bit weak here, I selected the plus or minus by comparing the floating point approx of the surd with that of the cosine. I couldn't think of a better way but if someone else can then please let me know. :)
 
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