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Simplifying a trigonometric expression

  1. May 31, 2008 #1
    cos(4x) + [tex]\sqrt{2}[/tex]cos(3x) + cos(2x) = 0
    ---------------------------------------------------

    Ok, so break down everything to a double angle and i get:

    2cos^2(2x) - 1 + sqrt(2)*(sqrt((cos(2x) + cos^2(2x))/2) - sqrt((1-cos^(2x)*(1-cos(2x))/2) + cos(2x) = 0

    ...which is quite complicated to solve even if i substitute cos(2x) with 't'.

    I also tried another aproach:

    2cos^2(2x) - 1 + sqrt(2)*(cos(2x)cos(x) - sin(2x)(sin(x)) + 2cos(2x) - 1 = 0

    here i aimed to complete a square but i cant see how.

    Can you please give me some advice on how to continue or use another method
     
  2. jcsd
  3. May 31, 2008 #2
    Hello.

    I used the Sum and Difference Formula for cosines.
    Def: cos(a)+cos(b) = 2*cos((a+b)/2)*cos((a-b)/2)

    Then, I let a=4x and b=2x

    The original equation's LHS becomes
    = 2cos((4x+2x)/2)*cos(4x-2x/2)) + [tex]\sqrt{2}[/tex]cos(3x)
    = 2cos(3x)cos(2x) + [tex]\sqrt{2}[/tex]cos(3x) by simplifying
    = cos(3x)[2cos(2x) + [tex]\sqrt{2}[/tex]] by distributive property
    Thus, cos(3x) = 0 or 2cos(2x) + [tex]\sqrt{2}[/tex] = 0

    Hope this helped.
     
    Last edited: May 31, 2008
  4. May 31, 2008 #3
    Thank you
     
  5. May 31, 2008 #4
    Didnt you make a mistake?
    Shouldn't it be = 2cos(3x)cos(x) + [tex]\sqrt{2}[/tex]cos(3x)?

    Anyway thanx for the help.
     
  6. May 31, 2008 #5
    Yes, sorry.
     
  7. Jun 2, 2008 #6
    sin(3x) = cos(x-30)
    How about this one?

    Where should I start from?
     
  8. Jun 2, 2008 #7
    Ok i got it :)
     
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