In answering this question, I needed to show that $\arctan \left( \frac{\tan \theta}{2} \right) \approx \frac{\theta}{2}$ when $\theta$ was small. So, naturally, I computed the first few terms of the Taylor series for $\arctan \left( \frac{\tan \theta}{2} \right)$. Sure enough, the first three terms are $0 + \frac12 \theta + 0 \theta^2$.
But what I didn't expect was the striking pattern in the next few terms: $$ \arctan \left( \frac{\tan \theta}{2} \right) = \frac12 \theta + \frac{1}{8} \theta^3 + \frac{1}{32} \theta^5 + \frac{11}{1920} \theta^7 + \cdots $$ It clearly seems to be the case that \begin{align*} \arctan \left( \frac{\tan \theta}{2} \right) &\approx \sum_{n \ge 0} \left(\frac{\theta}{2}\right)^{2n+1} \\ &= \frac{\theta / 2}{1 - \theta^2 / 4} \\ &= \frac{1}{2 - \theta} - \frac{1}{2 + \theta}. \\ \end{align*} Here is a plot:
Honestly, I have no idea how to explain this, and in particular I'm having trouble assigning any meaning to $\frac{1}{2 - x}$ at all when $x$ is an angle.
So, my question: is there any intuitive geometric reason why this approximation works? If not, is there any other concise reason?
Obviously one can arrive at this result via algebraic manipulation (like I did above) but I am interested in a more intuitive explanation.
