<s>[INST] How to approximate the value of $ \arctan(x)$ for $x> 1$ using Maclaurin's series? The expansion of $f(x) = \arctan(x)$ at $x=0$ seems to have interval of convergence $[-1, 1]$
$$\arctan(x) = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\mathcal{O}\left(x^{13}\right) $$
Does it mean that I cannot approximate $\arctan(2)$ using this series? Also I'm getting radius of convergence $|x| < 1$ using ratio test. How do I get $|x| \leqslant 1$?
 [/INST] Use, for $x>0$
$$
   \arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2}
$$
For $x<0$ use the parity.

Added: Use
$$
    \tan\left(\frac{\pi}{2} - \phi\right) = \frac{\sin\left(\frac{\pi}{2}-\phi\right)}{\cos\left(\frac{\pi}{2}-\phi\right) } = \frac{\cos(\phi)}{\sin(\phi)} = \frac{1}{\tan(\phi)}
$$
Thus, for $x = \tan(\phi)$:
$$
    \tan\left(\frac{\pi}{2} - \phi\right) = \frac{1}{x}, \quad \frac{\pi}{2} - \phi = \arctan\left(\frac{1}{x}\right), \quad 
   \frac{\pi}{2} - \arctan(x) = \arctan\left(\frac{1}{x}\right)
$$
</s>