TITLE: Why is this number so close to $1$?
QUESTION [12 upvotes]: The only positive solution of the equation $\sin (\tan x) = x$ is at a number $a = 0.999906...$. Is it a coincidence that the number $a$ is so close to $1$, or is there a conceptual explanation?
It was obvious $a$ was going to have to be less than $1$, but not this close. The answer seems to be related to $\tan 1 \approx \pi/2$, but that merely shifts the problem to explaining why that is the case. 
(I became interested in this question after reading this other question: Prove: $\sin (\tan x) \geq {x}$ )
Edit: I have thought about the question some more, and added some ideas in the form of an answer partially addressing why we have $\tan 1 \approx \pi/2$.

REPLY [5 votes]: J Swanson and Christian Blatter's answers address the question of why, given that $\tan 1 \approx \pi/2$ (within $1 \%$), we have $\sin \tan 1 \approx 1$ within $0.01\%$. Taylor's formula for $\sin x$ about $x = \pi/2$ shows why the accuracy would be approximately squared.
That leaves open the question of why $\tan 1 \approx \pi/2$. 
Gauss's continued fraction for $\tan z$ (see http://en.wikipedia.org/wiki/Gauss_continued_fraction ) is:
$$\tan z = \cfrac{z}{1 - \cfrac{z^2}{3 - \cfrac{z^2}{5 - \cfrac{z^2}{7 - {}\ddots}}}}.$$
In particular, for $z = 1$, this yields
$$ \tan 1 \approx  \cfrac{1}{1 - \cfrac{1}{3 - \cfrac{1}{5}}} = \frac{14}{9}.$$
This is an excellent approximation, valid to within just over $0.1\%$. So we need only consider the question of why $\pi \approx 28/9$. 
I don't have a very good answer for this part, as I haven't been able to locate any standard approximating sequences of $\pi$ one of whose first terms is $28/9$. However, you can obtain $28/9$ as an approximation of $\pi$ in the following way, as explained in A History of Pi by Beckmann. (If the circle has area $\pi$, then the octagon has area $28/9$.)