In a chapter of Computational Algebra and Number Theory called Continued Fractions of Algebraic Numbers (Available at http://citeseerx.ist.psu.edu/viewdoc/download?rep=rep1&type=pdf&doi=10.1.1.135.107) , Bombieri and van der Poorten give conditions under which the continued fraction of some algebraic numbers may be calculated by several complicated recursive formulas.
Experimentation has led me to believe that these same formulas can be used to calculate the continued fraction representations of at least some transcendental numbers as well, although a proof is lacking.
The theorem giving the formulas for the continued fraction expansions is Theorem 3 on page 151 of the paper by Bombieri and van der Poorten.
Here I'll give the recursive formulas as they were used by me in a program written in Mathematica.
k = 1;
f[x_] := Cos[ k x];
p[0] = 1; p[-1] = 1;
q[0] = 1; q[-1] = 0;
p[h_] := p[h] = c[h] p[h - 1] + p[h - 2];
q[h_] := q[h] = c[h] q[h - 1] + q[h - 2];
r[h_] := r[h] = p[h]/q[h];
c[h_] := c[h] =
Floor[((-1)^h ( f'[r[h - 1]])/((q[h - 1]^2) f[r[h - 1]])) -
q[h - 2]/q[h - 1] + ((-1)^(h - 1)/q[h - 1]^2)
Sum[(1/(r[h - 1] - (j + 1/ 2) Pi/k)) + (1/(r[h - 1] - (-j + 1/2) Pi/k)),
{j, 1, Infinity}]];
$$\begin{eqnarray}c(h+1) &=& \bigg\lfloor (-1)^{h+1} \frac{f'(r(h))}{q(h)^2 f(r(h))} - \frac{q(h-1)}{q(h)} \newline & &+ \frac{(-1)^h}{q(h)^2}\sum_{j=1}^\infty \left(\frac{1}{r(h) - (j+\frac{1}{2})\frac{\pi}{k}} + \frac{1}{r(h)-(-j+\frac{1}{2}) \frac{\pi}{k}} \right)\bigg\rfloor\end{eqnarray}$$
Letting the function f be cos(x) gives rise to the partial quotients for $2/(\pi-2)$. It is this that I'd like to prove.
