I am trying to approximate the expression
$\sin \varphi_s - \sin \varphi_e = 2 \cos\left(\frac{\varphi_s + \varphi_e}{2}\right) \sin\left(\frac{\varphi_s-\varphi_e}{2}\right)$
to include the angle $\theta_s = \arctan \left(\frac{x_m}{y_m}\right)$ .
Hereby an image of what I mean:
Of course we also know that
$\sin \varphi_s = \frac{x_m + \frac{L}{2}}{\sqrt{(x_m + \frac{L}{2})^2 + y_m^2}} = \frac{r_m \sin\theta_s + \frac{L}{2}}{\sqrt{(r_m\sin\theta_s + \frac{L}{2})^2 + r_m^2\cos^2\theta_s}} = \frac{r_m \sin\theta_s + \frac{L}{2}}{\sqrt{r_m^2 + L r_m \sin\theta_s + \frac{L^2}{4}}}$
and similarly
$\sin \varphi_e = \frac{x_m - \frac{L}{2}}{\sqrt{(x_m - \frac{L}{2})^2 + y_m^2}} = \frac{r_m \sin\theta_s - \frac{L}{2}}{\sqrt{(r_m\sin\theta_s - \frac{L}{2})^2 + r_m^2\cos^2\theta_s}} = \frac{r_m \sin\theta_s - \frac{L}{2}}{\sqrt{r_m^2 - L r_m \sin\theta_s + \frac{L^2}{4}}}$,
where $L$ is highlighted in red in the diagram. I know that a good approximation for $L \ll r_m \land \theta_s \approx 0$ is obviously
$\sin \varphi_s - \sin \varphi_e \approx \frac{L}{\sqrt{x_m^2 + y_m^2}} = \frac{L}{r_m}$.
I also know from playing around with some numerical examples that
$\sin \varphi_s - \sin \varphi_e \approx \frac{L \cos^2 \theta_s}{r_m}$
is a good approximation for $L \ll r_m$ and a wide range of angles $\theta_s$, however I have trouble proving it using the expressions/fractions shown above. I've tried approximating the resulting triangles by two right triangles with height $r_m$ and base given by a projection of $L$ but I can't seem to simplify the expressions far enough to be useful.
Is there a straightforward way of arriving at the $\cos^2 \theta_s$ expression?
If $\sin\theta_s$ is small, there is the standard expansion of the two terms in their Taylor series around $\sin \theta_s=0$: $$ \sin\varphi_s - \sin\varphi_e = \frac{r_m\sin\theta_s+L/2}{\sqrt{r_m^2+Lr_m\sin\theta_s+L^2/4}} -\frac{r_m\sin\theta_s-L/2}{\sqrt{r_m^2-Lr_m\sin\theta_s+L^2/4}} $$ $$ = \frac{2L}{\sqrt{4r_m^2+L^2}} - \frac{4r_m^2L(8r_m^2-L^2)}{(4r_m^2+L^2)^{5/2}}\sin^2\theta_s $$ $$ - \frac{20r_m^4L^3(16r_m^2-3L^2)}{(4r_m^2+L^2)^{9/2}}\sin^4\theta_s - \frac{168r_m^6L^5(24r_m^2-5L^2)}{(4r_m^2+L^2)^{13/2}}\sin^6\theta_s -\cdots $$ This is a sequence with a quite regular growth pattern in all the coefficients. The integers 4, 20, 168, 1716,... in the numerator are 4 times the Catalan numbers of https://oeis.org/A024492 . One can substitute $\sin^{2i}\theta_s=(1-\cos^2\theta_s)^i$ to obtain a series of $\cos\theta_s$.
Alternatively one can use a taylor series in terms of $L/r_m$: $$ \sin\varphi_s - \sin\varphi_e \approx \cos^2\theta_s (L/r_m) +\frac{1}{8}\cos^2\theta_s (5\sin^2\theta_s-1)(L/r_m)^3 $$ $$ +\frac{3}{128}\cos^2\theta_s (21\sin^4\theta_s-14\sin^2\theta_s+1)(L/r_m)^5 +O((L/r_m)^7) $$