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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?

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  • $\begingroup$ By $\delta_a-\delta_b\ll 1$, we have $\delta_a\approx\delta b$ and so $\cos\delta_a\cos\delta_b\approx \cos\delta_a\cos\delta_a$. $\endgroup$ Oct 11, 2023 at 14:44
  • $\begingroup$ How does that make sense? $ \delta_a $ could be 0.0009 and $ \delta_b $ 0.000009 and that is way less than 1 but they are 2 orders of magnitude apart, so not really close values! $\endgroup$
    – Jon
    Oct 11, 2023 at 15:28

1 Answer 1

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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.

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  • $\begingroup$ Is still quite hand-wavy, isn't it? One could go with the same argument and say that $ \sin(\delta_a - \delta_b) \approx \delta_a - \delta_b $ rather than 0 and get a completely different approximation. $\endgroup$
    – Jon
    Oct 13, 2023 at 10:24
  • $\begingroup$ For sure, but if you do so you'll find that the $\cos$ term is still doing most of the work. If we were doing this more formally then we would quantify the error of the approximation - for example, $\cos(x) = 1 + O(x^2)$, meaning that the error in approximating $\cos(x)$ as 1 is on a similar scale as $x^2$. $\endgroup$
    – ConMan
    Oct 14, 2023 at 15:02

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