# Deriving small angle approximation $\cos\delta_a\cos\delta_b\ \approx \cos^2\delta_a$

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?

• By $\delta_a-\delta_b\ll 1$, we have $\delta_a\approx\delta b$ and so $\cos\delta_a\cos\delta_b\approx \cos\delta_a\cos\delta_a$. Oct 11, 2023 at 14:44
• 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!
– Jon
Oct 11, 2023 at 15:28

$$\begin{eqnarray} \cos \delta_b & = & \cos (\delta_b - \delta_a + \delta_a) \\ & = & \cos (\delta_b - \delta_a) \cos \delta_a - \sin (\delta_b - \delta_a) \sin \delta_a \\ & = & \cos(\delta_a - \delta_b) \cos \delta_a + \sin(\delta_a - \delta_b) \sin \delta_a \end{eqnarray}$$
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.
• 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.
• 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 $x^2$. Oct 14, 2023 at 15:02