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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 $0\leqslant\zeta<1$.

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up vote 2 down vote accepted

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.

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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.

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