<s>[INST] Show that $\tan^{-1} (\tfrac{1}{8} ) \geq \frac{1}{\sqrt{65}} $ Here I'd like to check that:
$$ \frac{1}{8} \geq \tan^{-1} (\tfrac{1}{8} ) \geq \frac{1}{\sqrt{65}} $$
The numbers do check out if one uses a calculator.  They were using the Taylor series:


*

*$\frac{1}{\sqrt{65}} \;\;\;\;\;= 0.12403 $

*$\tan^{-1} \frac{1}{8} = 0.12435 $

*$\frac{1}{8} = 0.125$


The Taylor series for tangent could lead to an approximation. If $0 < x < \frac{\pi}{4}$ I believe we have:
$$ x - \frac{x^3}{3} < \tan^{-1} x <  x  $$
I do not recommend either of these, as a first recourse, but it came up in discussion last time.

These numbers might be somewhat mysterious (and they are). I got these using complex numbers.  I said that:
\begin{eqnarray*}
5 &=& 1^2 + 2^2 \\
13 &=& 2^2 + 3^2
\end{eqnarray*}
And there are angles associated to these sum of squares $\theta_5 = \tan^{-1} \frac{1}{2}$ and $\theta_{13} = \tan^{-1} \frac{2}{3} $
then, using the difference of arctangents formula:
$$ \tan^{-1} \frac{2}{3} - \tan^{-1} \frac{1}{2} = \tan^{-1} \left[
\frac{ \frac{2}{3} - \frac{1}{2}}{1 + \frac{2}{3} \times \frac{1}{2}} \right] = \tan^{-1} \tfrac{1}{8}$$
and this leads to the left side inequalty above.
 [/INST] Suppose $\theta = \tan^{-1}\frac18$. Then $\theta = \sin^{-1}\frac{1}{\sqrt{65}}$. Thus $\sin \theta = \frac{1}{\sqrt{65}}$. But $\theta > \sin\theta$.
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