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Find another expression form

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Hi,
    I need to prove that:
    x[t]=cos(ω0*t)+ cos( ωo*t + Δω*t)

    can be transformed into the form:
    x[t]=A(t)*cos[ωo*t + θ(t)]

    where A(t) and θ(t) are function of Δω.

    I have the solution but I cannot find out the way to solve it
    A(t)=2|cos(Δω*t)|

    and
    θ(t)= ArcTan[sin(Δω*t)/(1+cos(Δω*t))]

    here I can not figure out how to fin A(t) and θ(t).

    please can someone help me ?
    thank you
    B
    3. The attempt at a solution

    I have started by using the trigon identity cos(a+b) expansion.

    Then, I factor cos[ωo*t] to have 1+cos(Δω*t) and I factor 1+cos(Δω*t) to have the expression under the Arctan.

    OK I have:
    [tex]
    [1+\cos (\Delta \omega t) ] [\cos (\omega_0 t) - \sin (\omega_0 t)\frac{\sin (\Delta \omega t)}{1+\cos (\Delta \omega t)}]
    [/tex]

    Now let
    [tex]
    \theta(t)=\arctan(\frac{\sin (\Delta \omega t)}{1+\cos (\Delta \omega t)})
    [/tex]

    After that I am stuck..:confused:
    I dont know how to continue the transfromation to have another expansion od the type cos(a+b).
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 24, 2007 #2

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In addition to cos(a+b), can you use cos(a-b)?
     
  4. Oct 24, 2007 #3
    Yes I think I can.
     
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