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: 2,267 Thanks: 934 Math Focus: Wibbly wobbly timeywimey 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 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 
Senior Member Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402  
March 24th, 2019, 02:30 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,969 Thanks: 2219 
$0 = \cos^2(x) + \cos^2(2x) + \cos^2(3x)  1 = 2\cos(x)\cos(2x)\cos(3x)$, which is easy to solve.

March 24th, 2019, 05:39 PM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,267 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  
March 25th, 2019, 03:08 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,969 Thanks: 2219 
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. $\displaystyle \sin(x)\left(\sum_{k=1}^{4n1}\cos^2(kx)  (2n  1)\right) \equiv 2\sin(nx)\cos(nx)\cos(2nx)\cos((4n1)x)$ 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. 

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cosines, equation, trigonometry 
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