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Math Help - Angle sum.

  1. #1
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    Angle sum.

    How do I show that

    x(t)=\frac{\omega\cos(\omega t)+(1-\omega^{2})\sin(\omega t)}{\omega^{2}+(1-\omega^{2})^{2}}
    is the same as

    x(t)=K\sin\left[\omega\left(t-t_{0}\right)\right]

    where

    K=\frac{1}{\sqrt{\omega^{4}-\omega^{2}+1}}.
    Last edited by Cairo; March 4th 2010 at 01:34 PM.
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  2. #2
    Moo
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    Hello,

    The idea is to find a factor such that \omega \cos(\omega t)+(1-\omega^2)\sin(\omega t) can be written as a(\cos(\omega t)\sin(\theta)+\sin(\omega t)\cos(\theta)a \sin(\omega t+\theta) for some \theta

    So a has to satisfy : a^2=\omega^2+(1-\omega^2)^2=\omega^4-\omega^2+1 (because we have a\sin\theta=\omega and a\cos\theta=1-\omega^2)

    Hence \omega \cos(\omega t)+(1-\omega^2)\sin(\omega t)=\sqrt{\omega^4-\omega^2+1} \cdot \sin(\omega t-\theta)

    And then \frac{\omega \cos(\omega t)+(1-\omega^2)\sin(\omega t)}{\omega^2+(1-\omega^2)^2}=\frac{\sqrt{\omega^4-\omega^2+1} \cdot \sin(\omega t-\theta)}{\omega^4-\omega^2+1}

    Now let t_0 such that \theta=-\omega t_0 and you're done
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