By numerical computation it seems like, if $a_0 < a_1$: $$ \begin{multline} \log({a_0}^2 + {a_1}^2 + 2 a_0 a_1 \cos(\omega t)) = \log({a_0}^2 + {a_1}^2) \\ + \frac{a_0}{a_1}\cos(\omega t) - \frac{1}{2}\frac{{a_0}^2}{{a_1}^2}\cos(2\omega t) + \frac{1}{3}\frac{{a_0}^3}{{a_1}^3}\cos(3\omega t) - \frac{1}{4}\frac{{a_0}^4}{{a_1}^4}\cos(4\omega t) \ldots \end{multline} $$ If $a_0 > a_1$, the two term must be exchanged.

I'm quite confident in this solution but I cannot find a way to prove it mathematically...

Would be nice to know the reason of this result!