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Math Help - find a different form for trigonometric expression

  1. #1
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    find a different form for trigonometric expression

    Hi, I need to prove that:
    x[t]=cos(ω0*t)+ cos( ωo*t + Δω*t)

    can be transform 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))]

    I have started by using the trigon identity cos(a+b).
    I first factor cos[ωo*t] to have 1+cos(Δω*t) then I factor 1+cos(Δω*t) to have the expression under the Arctan.

    At one point I get:

    x[t]= (1+cosΔω*t)*[cos[ωo*t]-sin(ωo*t)*{sin(Δω*t)/(1+cosΔω*t)]}]

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

    please can someone help me to finsh the computation?
    thank you
    B
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  2. #2
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    Quote Originally Posted by braddy View Post
    Hi, I need to prove that:
    x[t]=cos(ω0*t)+ cos( ωo*t + Δω*t)
    If you have A\cos x + B\sin y where the coefficients are non-zero you can write,
    \sqrt{A^2+B^2} \left( \frac{A}{\sqrt{A^2+B^2}}\cos x + \frac{B}{\sqrt{A^2+B^2}}\sin x \right)
    If \theta = \tan^{-1} \frac{B}{A} then,
    \sqrt{A^2+B^2} \left( \cos x \cos \theta + \sin x \sin \theta \right) = \sqrt{A^2+B^2} \cos (x - \theta)

    Here expand,
    \cos (\omega_0 t) + \cos (\omega_0 t) \cos (\Delta \omega t) - \sin (\omega_0 t) \sin (\Delta \omega t)
    Thus,
    [ [1+\cos (\Delta \omega t) ] \cos (\omega_0 t) - \sin (\omega_0 t)
    Now apply the trick I used above.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    If you have A\cos x + B\sin y where the coefficients are non-zero you can write,
    \sqrt{A^2+B^2} \left( \frac{A}{\sqrt{A^2+B^2}}\cos x + \frac{B}{\sqrt{A^2+B^2}}\sin x \right)
    If \theta = \tan^{-1} \frac{B}{A} then,
    \sqrt{A^2+B^2} \left( \cos x \cos \theta + \sin x \sin \theta \right) = \sqrt{A^2+B^2} \cos (x - \theta)

    Here expand,
    \cos (\omega_0 t) + \cos (\omega_0 t) \cos (\Delta \omega t) - \sin (\omega_0 t) \sin (\Delta \omega t)
    Thus,
    [1+\cos (\Delta \omega t) ] \cos (\omega_0 t) - \sin (\omega_0 t)
    Now apply the trick I used above.
    Hi perfectHacker,
    I tried but I am still stuck. First
    [ [1+\cos (\Delta \omega t) ] \cos (\omega_0 t) - \sin (\omega_0 t)(You forgot the last term sin( delta omega*t))

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

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

    Now the all square root factor confuses me since I have not exactly the same form.
    Please can push just a little bit so I can finish it?
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  4. #4
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    (Yes I forgot that term).
    [1+\cos (\Delta \omega t) ] \cos (\omega_0 t) - <br />
\sin (\Delta \omega t)\sin (\omega_0 t)<br />
    So A = 1+\cos (\Delta \omega t) and B=\sin (\Delta \omega t).
    Thus,
    \sqrt{A^2+B^2} = \sqrt{ 1 + 2 \cos (\Delta \omega t) + \cos^2 (\Delta \omega t) + \sin (\Delta \omega t) }
    Simplify,
    \sqrt{2+2 \cos (\Delta \omega t)} = \sqrt{ 2(1+\cos (\Delta \omega t)}
    Half-angle idenity,
    \sqrt{4 \cos^2 \frac{\Delta \omega t}{2} } = 2<br />
\left| \cos \frac{\cos (\Delta \omega t)}{2} \right|

    (You can probably ignore absolute values because in this problem whatever you are modelling the angle is small enough so it remains positive).
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  5. #5
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    Thank you. Sorry I could not come earlier!!
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