December 18th, 2018, 11:43 AM  #11 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  
December 18th, 2018, 11:46 AM  #12 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra  is it correct? 
December 18th, 2018, 11:54 AM  #13 
Senior Member Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337  
December 18th, 2018, 01:00 PM  #14 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra 
The first one is all clear. I made my way through it, but can you clarify that second one more for me?
Last edited by skipjack; December 18th, 2018 at 01:45 PM. 
December 18th, 2018, 01:54 PM  #15 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
Expanding 1  (1  cos(x)) (4cos²(x) + 2cos(x)  1)² gives 16cos$^5\!$(x)  20cos³(x) + 5cos(x).

December 18th, 2018, 02:11 PM  #16 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra 
Thank you. What about deduce that cos (2pi / 5) is a solution of equation (E): 4X²  2X  1 = 0?
Last edited by skipjack; December 18th, 2018 at 02:31 PM. 
December 18th, 2018, 02:36 PM  #17 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
I think you meant 4X² + 2X  1 = 0. The second equation gives (1  cos(x)) (4cos²(x) + 2cos(x)  1)² = 1  cos(5x). Substituting x = 2$\pi$/5 gives (1  cos(2$\pi$/5))(4cos²(x) + 2cos(x)  1)² = 0. Is that sufficient help? 

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