Which is bigger $\tan^{-1}(\frac{1}{2}) $ or $\frac{1}{\sqrt{5}}$? A number theory textbook asked us to compare $\tan^{-1}(\frac{1}{2})$ and $\sqrt{5}$.  In fact, these are rather close:

\begin{eqnarray*}
\tan^{-1} \frac{1}{2} &amp;=&amp; 0.46364 \\ \\
\frac{1}{\sqrt{5}} &amp;=&amp; 0.44721
\end{eqnarray*}

So at least numerically I think we have the answer that the first one is bigger.  Momentarily, I thought we had an exact answer: $\tan^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4} $, but that's totally different.  So we are left with:

$$ \tan^{-1} \frac{1}{2}  &gt; \frac{1}{\sqrt{5}} &gt; 0 $$

It's impressive that we could have so many decimal places, and I wonder if I should take the computer on faith for that.  And I noticed these two answers are close, so I also wonder if we estimate the difference... I don't have any conjecture in either case yet.

For now I just want proof of the inequality as stated above.
Suppose $\tan(x)=\frac{1}{2}$. Then in particular you know $\frac{\sin(x)}{\cos(x)}=\frac{1}{2}$. You also know $\sin^2(x)+\cos^2(x)=1$. Using the last two equalities you find that $\sin(x)=\frac{1}{\sqrt{5}}$. Since $x\geq\sin(x)$, you're done.
This seems like more of a real analysis question than number theory. The first method I thought of was taking the Taylor series for tangent($\sqrt{5}$) and comparing that to $\frac{1}{2}$ (and noting that tangent is an increasing function), but arctan has a nicer Taylor series than tangent does. Either way, what you can do is take the first n Taylor series terms, argue that the remaining terms add up to no more than some error term, and if you've taken a large enough n, then the error term will be small enough that you can say which is larger, even with the error term. That's the general real analysis strategy for comparing number obtained from analytic functions. There might be some clever number theory method I'm missing.