1. ## 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

$K=\frac{1}{\sqrt{\omega^{4}-\omega^{2}+1}}.$

2. 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