1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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 :)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook