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!



Fourier series of $\log(a +b\cos(x))$?
Let's consider
$$
f(x) = \log (1+q^2+2q\cos x) = \log |1+qe^{ix}|^2 ,
$$
which differs from your function only by the additive constant $2\log a\_1$ if we take $q=a\_0/a\_1$. Since $|q|<1$, we can use the Taylor series of $\log(1+z)$ to write
$$
\begin{align}
f(x) = 2\,\textrm{Re}\; \log (1+qe^{ix}) = 2\,\textrm{Re}\sum\_{n\ge 1} (-1)^{n-1}\frac{q^n}{n} e^{inx} \\
= 2\sum\_{n\ge 1} (-1)^{n-1}\frac{q^n}{n} \cos nx .
\end{align}
$$


