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 February 3rd, 2015, 06:23 AM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Roots of unity. Find the fifth roots of unity. If $\displaystyle \omega$ is the root with smallest positive argument if $\displaystyle u=\omega +\omega^4$ and $\displaystyle v=\omega^2+\omega^3$ Show that $\displaystyle u+v=-1$ and that $\displaystyle u-v=sqrt5$ Hence find $\displaystyle \cos72^{\circ}$
 February 3rd, 2015, 06:54 AM #2 Member   Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10 here's the first part: w=e^(i2pi/5) u+v is the sum of the 5th primitive roots of unity, which is (w^5-1)/(w-1)-1. e^(i2pi/5)^5=e^(i2pi)=0 so we are left with -1 Thanks from topsquark
 February 3rd, 2015, 07:20 AM #3 Member   Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10 To find cos(72), note that cos(72)+cos(144)+cos(216)+cos(288 )=-1, and then spam the double angle identity until you're left with only cos(72), then you can solve the polynomial for cos(72). After that the second part (u-v) should be easy Thanks from topsquark
February 3rd, 2015, 07:36 AM   #4
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Quote:
 Originally Posted by USAMO Reaper To find cos(72), note that cos(72)+cos(144)+cos(216)+cos(288 )=-1, and then spam the double angle identity until you're left with only cos(72), then you can solve the polynomial for cos(72).
Code:
algdep(cos(72*Pi/180),2)

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### buktikan bahwa 1 cos 72 cos 144 cos 216 cos 288=0

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