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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 $\displaystyle uv=sqrt5$ 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^51)/(w1)1. e^(i2pi/5)^5=e^(i2pi)=0 so we are left with 1 
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 (uv) should be easy

February 3rd, 2015, 07:36 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  

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