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December 18th, 2018, 09:34 AM   #1
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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
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December 18th, 2018, 09:48 AM   #2
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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).
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December 18th, 2018, 09:57 AM   #3
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Unfortunately, I'm not, I'm still not quite there yet.

Last edited by skipjack; December 18th, 2018 at 10:30 AM.
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December 18th, 2018, 10:01 AM   #4
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Quote:
Originally Posted by SkyCod View Post
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

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.
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December 18th, 2018, 10:06 AM   #5
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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.
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December 18th, 2018, 11:20 AM   #6
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Thing about splitting cos(5x) into cos(3x + 2x)
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December 18th, 2018, 11:24 AM   #7
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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)
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December 18th, 2018, 11:28 AM   #8
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Quote:
Originally Posted by skipjack View Post
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.
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December 18th, 2018, 11:35 AM   #9
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Quote:
Originally Posted by romsek View Post
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
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December 18th, 2018, 11:38 AM   #10
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Quote:
Originally Posted by SkyCod View Post

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
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