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Thread: Angle sum.

  1. #1
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    Angle sum.

    How do I show that

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

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

    where

    $\displaystyle K=\frac{1}{\sqrt{\omega^{4}-\omega^{2}+1}}.$
    Last edited by Cairo; Mar 4th 2010 at 12:34 PM.
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  2. #2
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    Hello,

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

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

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

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

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