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 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 $(\cos(x)+i \sin(x))^5 = \sum \limits_{k=0}^5~\dbinom{5}{k}\cos^k(x) (i \sin(x))^{5-k}$ $\cos(5x) = Re\left(\sum \limits_{k=0}^5~\dbinom{5}{k}\cos^k(x) (i \sin(x))^{5-k}\right)$ I'll let you take it from there. The rest is just complex algebra (complex as in complex numbers). Thanks from topsquark and SkyCod 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
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Quote:
 Originally Posted by SkyCod unfortunetely im not , i'm still quiet not there yet.
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

$\cos(5x) = \cos(x+4x) = \cos(x)\cos(4x) - \sin(x)\sin(4x)$

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) Thanks from topsquark and SkyCod December 18th, 2018, 11:28 AM   #8
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Quote:
 Originally Posted by skipjack 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)
Thanks, any idea about the second one?

Last edited by skipjack; December 18th, 2018 at 11:43 AM. December 18th, 2018, 11:35 AM   #9
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Quote:
 Originally Posted by romsek 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. Typo on line 1:
$\displaystyle \cos(a + b) = cos(a)~cos(b) - sin(a)~sin(b)$

but you corrected for it later.

-Dan December 18th, 2018, 11:38 AM   #10
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Quote:
 Originally Posted by SkyCod cos(5x) = (1-cos(x)) (4cos²(x) + 2 cos(x) -1)² some help, please...
Go ahead and try it using the methods skipjack and romsek demonstrated for you. If you can't get it, please show us your work and we'll see where the error is.

-Dan Tags demonstration, easy, showing Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Carl James Mesaros Trigonometry 1 November 7th, 2017 09:27 AM Fernando89 Pre-Calculus 5 June 27th, 2017 07:11 PM AlbertoF Algebra 2 December 5th, 2015 02:31 PM Fingolfin Math Events 10 August 26th, 2012 08:21 AM Arhimede Number Theory 2 August 28th, 2009 07:45 AM

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