December 18th, 2018, 09:34 AM  #1 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra  Easy demonstration
Show that: cos(5x) = cos(x)(16 cos$^4$(x)  20 cos²(x) + 5) cos(5x) = 1  (1  cos(x)) (4cos²(x) + 2 cos(x)  1)² some help, please... Last edited by skipjack; December 18th, 2018 at 11:40 AM. Reason: to correct 2nd identity 
December 18th, 2018, 09:48 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,584 Thanks: 1430 
Are you familiar with complex numbers? The easiest way to show this is to let $z = e^{i x} = \cos(x)+i \sin(x)$ $z^5 = e^{i 5x} = \cos(5x) + i \sin(5x)= (\cos(x)+i \sin(x))^5$ $\cos(5x) = Re(z^5)$ use the binomial theorem to expand the right hand side $ (\cos(x)+i \sin(x))^5 = \sum \limits_{k=0}^5~\dbinom{5}{k}\cos^k(x) (i \sin(x))^{5k}$ $\cos(5x) = Re\left(\sum \limits_{k=0}^5~\dbinom{5}{k}\cos^k(x) (i \sin(x))^{5k}\right)$ I'll let you take it from there. The rest is just complex algebra (complex as in complex numbers). 
December 18th, 2018, 09:57 AM  #3 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra 
Unfortunately, I'm not, I'm still not quite there yet.
Last edited by skipjack; December 18th, 2018 at 10:30 AM. 
December 18th, 2018, 10:01 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,584 Thanks: 1430  well it's going to have to be brute force then the formulas $\cos(a+b) = \cos(a)\sin(b)  \cos(b)\sin(a)$ and $\cos(2x) = 2\cos^2(x)1$ and of course $\sin^2(x) = 1\cos^2(x)$ will come in handy I'd start with $\cos(5x) = \cos(x+4x) = \cos(x)\cos(4x)  \sin(x)\sin(4x)$ then $\cos(4x) = \cos(2(2x)) = 2 \cos^2(2x) 1$ and just keep chipping away at it until everything is converted to powers of $\cos(x)$ no magic, just a bunch of algebra to slog through applying these formulas over and over. 
December 18th, 2018, 10:06 AM  #5 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra 
Okay, I'll start digging. If anyone can do them, you're welcome to share. It will help.
Last edited by skipjack; December 18th, 2018 at 10:31 AM. 
December 18th, 2018, 11:20 AM  #6 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra 
Thing about splitting cos(5x) into cos(3x + 2x)

December 18th, 2018, 11:24 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 21,027 Thanks: 2258 
cos(2x) = 2cos²(x)  1 sin(2x) = 2sin(x)cos(x) sin²(x) = 1  cos²(x) cos(3x) = 4cos³(x)  3cos(x) sin(3x) = 3sin(x)  4sin³(x) = sin(x)(3  4sin²(x)) = sin(x)(4cos²(x)  1) cos(5x) = cos(3x + 2x) = cos(3x)cos(2x)  sin(3x)sin(2x) cos(5x) = (4cos³(x)  3cos(x))(2cos²(x)  1)  sin(x)(4cos²(x)  1)(2sin(x)cos(x)) cos(5x) = cos(x)((4cos²(x)  3)(2cos²(x)  1)  2(1  cos²(x))(4cos²(x)  1)) cos(5x) = cos(x)(16cos$^4\!$(x)  20cos²(x) + 5) 
December 18th, 2018, 11:28 AM  #8  
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra  Quote:
Last edited by skipjack; December 18th, 2018 at 11:43 AM.  
December 18th, 2018, 11:35 AM  #9  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \cos(a + b) = cos(a)~cos(b)  sin(a)~sin(b)$ but you corrected for it later. Dan  
December 18th, 2018, 11:38 AM  #10 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timeywimey stuff.  

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