My Math Forum Equation with cosines

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 March 24th, 2019, 08:10 AM #1 Newbie   Joined: Mar 2019 From: Europe Posts: 2 Thanks: 0 Equation with cosines Here is the equation. I tried using substitution and expanding the terms with the help of trigonometric identities but nothing has worked for me. Could you help me by solving it? Thank you very much. Last edited by rixxcz; March 24th, 2019 at 08:22 AM.
 March 24th, 2019, 08:57 AM #2 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,205 Thanks: 901 Math Focus: Wibbly wobbly timey-wimey stuff. Here are some highlights: $\displaystyle \cos(2x) = 2 \cos^2(x) - 1$ so $\displaystyle \cos^2(2x) = 4 \cos^4(x) - 4 \cos^2(x) + 1$ $\displaystyle \cos(3x) = \cos(x)~\cos(2x) - \sin(x)~\sin(2x) = \cos(x)(2 \cos^2(x) - 1) - \sin(x) ( 2 \sin(x)~\cos(x) )$ $\displaystyle \text{... } = 4 \cos^3(x) - 3 \cos(x)$ so $\displaystyle \cos^2(3x) = 16 \cos^6(x) - 24 \cos^4(x) + 9 \cos^2(x)$ Adding these together and putting them into the original equation: $\displaystyle \cos^2(x) + \cos^2(2x) + \cos^2(3x) = 1$ $\displaystyle 16 \cos^6(x) - 20 \cos^4(x) + 6 \cos^2(x) + 1 = 1$ Can you finish from here? (It's not nearly as bad as it looks.) -Dan Thanks from romsek and TheReluctantMathMan Last edited by skipjack; March 24th, 2019 at 01:52 PM.
 March 24th, 2019, 09:44 AM #3 Newbie   Joined: Mar 2019 From: Europe Posts: 2 Thanks: 0 Yes, I see it now. I just have to substitute $\displaystyle \cos^2(x) = u$ and solve the given equation as usual. Thank you Dan. Last edited by skipjack; March 24th, 2019 at 01:52 PM.
March 24th, 2019, 10:57 AM   #4
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Quote:
 Originally Posted by topsquark (It's not nearly as bad as it looks.) -Dan
I dunno... it looks pretty bad...

 March 24th, 2019, 02:30 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 $0 = \cos^2(x) + \cos^2(2x) + \cos^2(3x) - 1 = 2\cos(x)\cos(2x)\cos(3x)$, which is easy to solve.
March 24th, 2019, 05:39 PM   #6
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Quote:
 Originally Posted by skipjack $0 = \cos^2(x) + \cos^2(2x) + \cos^2(3x) - 1 = 2\cos(x)\cos(2x)\cos(3x)$, which is easy to solve.
Is that part of a general theorem or does it just work with this particular problem?

-Dan

 March 25th, 2019, 03:08 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 The identity I used is easy to prove by use of standard trigonometric identities, but can be obtained from a general result, as shown below. $\displaystyle \sin(x)\left(\sum_{k=1}^{4n-1}\cos^2(kx) - (2n - 1)\right) \equiv 2\sin(nx)\cos(nx)\cos(2nx)\cos((4n-1)x)$ For $n = 1$, that gives $\sin(x)(\cos^2(x) + \cos^2(2x) + \cos^2(3x) - 1) \equiv 2\sin(x)\cos(x)\cos(2x)\cos(3x)$, which implies the identity I used. For $n > 1$, the above generalization would usually be given in a simpler form. Thanks from topsquark

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