How to deal with approximations involving "large" numbers I came across this question:

Here I am given a circle, with center O.
2 points $S_1$ and $S_2$ are present on the diameter of the circle at equal distances from the center.
$P$ is a point on the circle.
I had to find the relation between the angle $\theta$ made by $OP$ with the line joining $S_1$ and $S_2$, the distance $S_1S_2$, and the difference between $PS_1$ and $PS_2$
But with a catch. The radius of the circle is much greater than $S_1S_2$.
So I proceeded with the following approximation: the angle made by $PS_2$ with $S_1S_2$ to be approximately equal to $\theta$.
And the difference between $PS_2$ and $PS_1$ to be the component of $S_1S_2$ along $PS_1$.

This gave me the difference between $PS_2$ and $PS_1$ as $S_1S_2\cos(\theta)$
I plugged this into my calculations and I got the answer!
However for some reason I decided to try it in a different way.
This time, seeing as I could approximate the angle between $PS_2$ and $S_1S_2$ as $\theta$ I approximated both the angles made by $PS_1$ and $PS_2$ with $S_1S_2$ as $\theta$.
Then drawing the perpendicular from $P$ to the diameter containing $S_1S_2$ at $M$, I evaluated $PS_1$ as $S_1M/\cos(\theta)$
And $PS_2$ as $S_2M/\cos(\theta)$.
Subtracting these I get the difference between $PS_2$ and $PS_1$ as $(S_2M-S_1M)/\cos(\theta)$.
Thus I got the difference between $PS_2$ and $PS_1$ as $(S_1S_2)/\cos(\theta)$.

This was totally different from my previous answer.
It seems to me that all these approximations are valid as the radius of the circle was much greater than $S_1S_2$.
I have spent a long time trying to understand what went wrong. None of my friends could find the error too.
I would be really glad if someone could point out why some approximations I took were not valid.
I would appreciate help in this regard.
 A: Here's a way to visualize the error in your second approximation.
The error is that pretending $\angle PS_1M = \angle PS_2M$ (because the angles are nearly equal) does not actually make these angles equal.
If we actually put a line through $S_1$ at the same angle $\theta$ as the line $S_2P,$ the parallel line through $S_1$ intersects $PM$ at $N$ (not $P$) as shown in the figure below.

If the angle $\theta$ is very small (nearly zero), the difference between the lengths $S_1N$ and $S_1P$ is not much; and indeed for very small $\theta,$ we have
$\cos\theta \approx 1$ and so
$$S_1S_2 \cos\theta \approx S_1S_2 \approx \frac{S_1S_2}{\cos\theta}. $$
The second approximation is still overestimates $S_2P - S_1P$
(mainly because of the error introduced by $S_1N < S_1P$),
but not by nearly as much as when $\theta$ is a larger acute angle.
When things get really bad is when $\theta$ gets closer to a right angle.
Then $S_1P - S_1N,$ the error you introduced by assuming non-parallel lines were parallel, gets closer to the distance $NP.$
And we can compute that $NP = S_1S_2/\cos\theta$ exactly,
that is, for $\theta$ near a right angle your approximation is really just finding the length of $NP$ and has practically nothing to do with
$S_2P - S_1P.$

In general, the assumption should be that an approximation does not work
until is is proven that it does work.
So the question really should be why the first approximation is any good.
The proof is that you constructed a small right triangle
$\triangle S_1QS_2$ whose right angle vertex $Q$ is on $S_2P,$ and therefore
$\triangle S_1QP$ also is a right triangle.
Moreover you can show that under the assumption that $S_1P$ is much greater than $S_1S_2$ (and therefore also much greater than $S_1Q$),
the difference between $S_1P$ and $QP$ is much less than $S_1Q$
and therefore much less than $S_1S_2.$
Since your approximation actually measures the difference
$S_2P - QP$ rather than $S_2P - S_1P,$
it introduces an error equal to $S_1P - QP,$ but this error is
small relative to the correct value.
A: In the second approach, we can't pretend that $\angle PS_1M$ and $PS_2M$ are exactly equal, so let $\angle PS_2M = \theta,$ and $\angle S_1PS_2 = \varepsilon,$ where $\varepsilon$ is "small", so
$$
\angle PS_1M = \theta + \varepsilon.
$$
Because $\varepsilon$ is "small", we have $\cos\varepsilon \bumpeq 1,$ with an error of order $\varepsilon^2,$ which we neglect, and so
$$
\cos(\theta + \varepsilon) \bumpeq \cos\theta - \sin\varepsilon\sin\theta.
$$
Dropping a perpendicular from $S_1$ to $PS_2,$ as in your second diagram, we have
$$
PS_1 \sin\varepsilon = S_1S_2\sin\theta.
$$
Therefore
\begin{align*}
PS_2 - PS_1 & = \frac{S_2M}{\cos\theta} - \frac{S_1M}{\cos(\theta + \varepsilon)} \\
& \bumpeq \frac{S_2M}{\cos\theta} - \frac{S_1M}{\cos\theta - \sin\varepsilon\sin\theta} \\
& = \frac{S_2M}{\cos\theta} - \frac{S_1M}{\cos\theta}(1  - \sin\varepsilon\tan\theta)^{-1} \\
& \bumpeq \frac{S_2M}{\cos\theta} - \frac{S_1M}{\cos\theta}(1  + \sin\varepsilon\tan\theta) \\
& = \frac{S_1S_2}{\cos\theta} - \frac{S_1M\sin\varepsilon\tan\theta}{\cos\theta} \\
& \bumpeq \frac{S_1S_2}{\cos\theta} - \frac{S_1M\sin\varepsilon\tan\theta}{\cos(\theta + \varepsilon)} \\
& = \frac{S_1S_2}{\cos\theta} - PS_1\sin\varepsilon\tan\theta \\
& = \frac{S_1S_2}{\cos\theta} - \frac{S_1S_2\sin^2\theta}{\cos\theta} \\
& = S_1S_2\cos\theta,
\end{align*}
in agreement with the first approach.
I expect this argument can be shortened.
