A relation between sine and cosine I cannot figure out how this relation:
$$\cos(\omega t)+ \frac{\zeta}{\sqrt{1-\zeta^2}}\sin(\omega t)  $$
is equal to:
$$\frac{1}{\sqrt{1-\zeta^2}}\sin\left(\omega t + \tan^{-1}\frac{\sqrt{1-\zeta^2}}{\zeta}\right)$$
I only found that this is not true for $\zeta<0$, and specifically it must be $0\leqslant\zeta<1$.

This probably comes from $a\sin(\gamma)+b\cos(\gamma)$. To simplify this, a key idea is to find a triangle with catets $a$ and $b$ and hypothenuse $\sqrt{a^2+b^2}$. You shall choose an angle $\theta$ from the triangle such that $\sin(\theta)=b/\sqrt{a^2+b^2}$ and $\cos(\theta)=a/\sqrt{a^2+b^2}$ so you can write $$a\sin(\gamma)+b\cos(\gamma)=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin(\gamma)+\frac{b}{\sqrt{a^2+b^2}}\cos(\gamma)\right)\\=\sqrt{a^2+b^2}\left(\cos(\theta)\sin(\gamma)+\sin(\theta)\cos(\gamma)\right)\\= \sqrt{a^2+b^2}\sin(\gamma+\theta).$$ Now, you can write $\theta$ using the relations from the triangle we wrote.
