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December 18th, 2018, 11:43 AM   #11
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Quote:
 Originally Posted by SkyCod . . . any idea about the second one?
I've corrected the second one for you. It's now easy to prove it from the first one.

 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
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Quote:
 Originally Posted by romsek $\cos(a+b) = \cos(a)\sin(b) - \cos(b)\sin(a)$
bah...

$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$

 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: 21,027 Thanks: 2258 Expanding 1 - (1 - cos(x)) (4cos²(x) + 2cos(x) - 1)² gives 16cos$^5\!$(x) - 20cos³(x) + 5cos(x). Thanks from SkyCod
 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: 21,027 Thanks: 2258 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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