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Linearizing a cosine

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    I make the product [tex]\cos(x)\cos(5x)[/tex] a sum of two cosines. I think this is called linearizing but please correct me if I'm wrong.
    (I need it to find the n-th derivative of that function).

    2. Relevant equations



    3. The attempt at a solution
    I know the answer is:
    [tex]\frac{\cos(4x)+\cos(6x)}{2}[/tex]
    - Thanks HP 50g ;) -
    But how do I get to this result?
    I think I should be using Moivre's formula but I get both sines and cosines in my formula.
     
  2. jcsd
  3. Nov 15, 2008 #2

    Hurkyl

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    The most straightforward way is to apply the basic trig identities. The relevant one is on the list of ones you should, if you intend to do a lot of arithmetic with trig functions, either have memorized, or know how to derive quickly.

    Did you finish simplifying the expression you got? What did you actually get?
     
  4. Nov 15, 2008 #3
    Well, I did get to this with Moivre's formula:
    [tex]16\cos^6 x-20\cos^4 x+5\cos^2 x[/tex]

    Obviously I could expand cos(4x) and cos(6x) and be like "Oh! If you sum them and divide by 2, you get that too!" but I would never think about expanding cos(4x) and cos(6x) if I didn't know the result beforehand.
     
  5. Nov 15, 2008 #4

    Hurkyl

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    What were your first couple steps?

    If you did what I think you did... you might consider not expanding your cis's, preferring to keep them in factored form.


    (If you haven't seen that abbreviation, the function cis is defined as [itex]cis(z) = cos(z) + i sin(z)[/itex])
     
  6. Nov 15, 2008 #5

    Hurkyl

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    Actually, you might! If you knew that the expansion of [itex]cos(nx)[/itex] has [itex]cos^n x[/itex] in it, you would consider expanding [itex]cos(6x)[/itex] to kill off the leading term, and then continue going from there.
     
  7. Nov 15, 2008 #6
    What I did:
    [itex]cis(z) = cis(nz) cos(nz) + i sin(nz) = [cos(z) + i sin(z)]^n[/itex]
    Then take real part. Then replace: [itex]sin^2 (x) = 1- cos^2 (x)[/itex]
    I don't really understand what you mean by not expanding my cis(z).

    Edit: That would leave me with something to the fourth power and I would think "maybe now try cos(4x)" and voilà! Yep... But I still would like to hear the explanation for cis(z) if it's possible.
     
  8. Nov 15, 2008 #7

    Hurkyl

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    I had assumed you started by doing something like

    [tex]
    \cos x \cos 5x = \frac{cis(x) + cis(-x)}{2} \frac{cis(5x) + cis(-5x)}{2}
    [/tex]

    or maybe

    [tex]
    \cos x \cos 5x = \mathcal{R}[ cis(x) ] \mathcal{R}[ cis(5x) ]
    [/tex]

    (Grrr, why can't I figure out how to typeset this right? :frown:)
     
    Last edited: Nov 15, 2008
  9. Nov 15, 2008 #8
    It was the latter. The former uses formulae I have never encountered.
     
  10. Nov 15, 2008 #9

    Hurkyl

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    Do you know how to write the "real part" function in terms of complex conjugation?
     
  11. Nov 15, 2008 #10
    [tex]\Re (z) = \frac{z + \overline{z}}{2}[/tex]
    Right?
     
  12. Nov 15, 2008 #11

    Hurkyl

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    Right. That's where the other formula I wrote came from.

    (Incidentally, similar ideas can be used to kill off unwanted terms in various trig identities...)
     
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