# Why does $\arctan(\frac{\tan \theta}{2}) \approx \frac{1}{2 - \theta} - \frac{1}{2 + \theta}$ for small $\theta$?

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.

• Go one more step, Neglect $\theta^2 <<1$ and you get there. Mar 1, 2018 at 21:25

The approximation you noticed is the Padé approximant of order $(m,n) = (1,2)$ of $f(\theta) = \tan^{-1} \bigl( \frac{1}{2} \tan \theta \bigr)$.
• Geometric means arising from a geometric picture. For instance there are very straightforward proofs that $\sin x / x$ tends to $1$ or that the area of a circle is $\pi r^2$ using geometry. And "intuitive" means you understand why it is, not just how to prove it.