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 $0\leqslant\zeta<1$.

The key thing to notice is that a right angled triangle with sides of length $\sqrt{1-\zeta^2}$ and $\zeta$ has hypotenuse of length $1$ by Pythagoras' Theorem. So 
$\sin(\tan^{-1}(\frac{\sqrt{1-\zeta^2}}{\zeta}) = \sqrt{1-\zeta^2}$ and 
$\cos(\tan^{-1}(\frac{\sqrt{1-\zeta^2}}{\zeta}) = \zeta$
Then use the rule for $\sin(a+b)$ and it should come out.
