I'm trying to understand the derivation for the small angular distance approximation formula for the Angular Distance, as explained in Wikipedia
Almost to the end of the derivation, I'm presented with a small angle approximation I can't make sense of:
Given that $ \delta_a - \delta_b \ll 1 $ and $ \alpha_a - \alpha_b \ll 1 $, at a second-order development it turns that
$$ \cos\delta_a\cos\delta_b\frac{(\alpha_a-\alpha_b)^2}{2} \approx \cos^2\delta_a\frac{(\alpha_a-\alpha_b)^2}{2} $$
This seems to imply that:
$$ \cos\delta_a\cos\delta_b\ \approx \cos^2\delta_a $$
How did they get this approximation?
We can expand things out using a standard "sum of angle" formula:
$$\begin{eqnarray} \cos \delta_b & = & \cos (\delta_b - \delta_a + \delta_a) \\ & = & \cos (\delta_b - \delta_a) \cos \delta_a - \sin (\delta_b - \delta_a) \sin \delta_a \\ & = & \cos(\delta_a - \delta_b) \cos \delta_a + \sin(\delta_a - \delta_b) \sin \delta_a \end{eqnarray}$$
Now, if $\delta_a - \delta_b \ll 1$, then $\cos(\delta_a - \delta_b) \approx 1$ and $\sin(\delta_a - \delta_b) \approx 0$, which reduces the above to $\cos \delta_b \approx \cos \delta_a$.
To use your example values from the comments, if $\delta_a = 0.0009$ and $\delta_b = 0.000009$, we have $\cos \delta_a \approx 0.9999999998766$ and $\cos \delta_b \approx 0.99999999999998766$, so the difference between the two is about $1.2 \times 10^{-10}$. Notice that this mostly comes about because both angles are so small that in both cases $\cos \delta \approx 1$ is already a reasonable approximation.