In Mathematical Astronomy Morsels, Meeus defines: $$ A = 60 + 2 \cos t +0.3 \cos 2t $$ And gives the following expression for calculating the inverse of $A$
1/A = 0.016 676 073 03
– 0.000 553 255 76 cos t
– 0.000 074 203 29 cos 2t
+ 0.000 002 617 46 cos 3t
+ 0.000 000 141 96 cos 4t
– 0.000 000 008 85 cos 5t
– 0.000 000 000 19 cos 6t
I did a Fourier cosine series expansion of $1/A$, the coefficients of which are given by: $$ C_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{dt}{60 + 2 \cos t +0.3 \cos 2t} \\ C_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{\cos nt}{60 + 2 \cos t +0.3 \cos 2t} dt $$ Evaluating the integrals numerically for $ n=1, 2, 3, \ldots 6$, I get the following for the coefficients
0.016 676 073 104 944
– 0.000 553 255 784 230
– 0.000 074 203 416 170
+ 0.000 002 617 518 128
+ 0.000 000 142 032 216
– 0.000 000 008 907 612
– 0.000 000 000 207 049
Problem is, from like 10th decimal place onwards my coefficients and the ones given by Meeus are different (mine are always a hair larger than Meeus).
So why don't the coefficients match?
To check your work you can get exact results by some purely algebraic methods. Introduce the symbol $X= \cos t$ and note that $A$ is a quadratic in $X$:
$A= 60 + 2 X +.30 ( 2X^2-1)$
The quadratic has roots $r_1,r_2= 1/6 (-10 \pm Sqrt[118])$ (I used Mathematica for that, but you should double-check it). Next you can expand $1/A$ by partial fractions then expand each of the two partial fraction terms as a Taylor series in $X$. Finally you can represent each resulting power of $X$ in these expansions, such as $X^8= \cos ^8 t$ as an explicit trigonometrical polynomial (finite trig series) that has integer coefficients. I suggest that you read about the Chebyshev polynomials for details of that last step. This is a tediously direct method, but it introduces no floating-point arithmetic errors.