Halving a tiny angle by doubling a side of a triangle. You have a triangle $ABC$ with a right angle $\angle CAB$. The line stretch between $A$ and $B$ shall be very long in comparison to the other sides. Now you are going to measure the angle $\angle ABC$ which is very tiny, e.g. $1'$. 
Now you keep the line stretch between $A$ and $C$ constant. Next you double the length of $AB$ and $BC$ is adapted according to the Pythagorean theorem. 
According to my physiology textbook: If $\angle ABC$ is tiny and I am going to double $AB$ as above, $\angle ABC$ should be halved. 
How to prove that?

The rough idea is as Vizzy says: for small angles $\theta$, $\tan \theta \approx \theta$.
We can make this rigorous.
Suppose the original angle is $\theta$, and that originally $AC = l$ and $AB = L$
($L \gg l$).
Then
$$
\tan \theta = \frac{l}{L}
$$
so after doubling the length of $AB$, the new angle $\theta_1$ will satisfy
$$
\tan \theta_1 = \frac{l}{2L} = \frac{\tan \theta}{2}
$$
Therefore,
$$
\theta_1 = \arctan \left( \frac{\tan \theta}{2} \right).
$$
Let $f(x) = \arctan \left( \frac{\tan x}{2} \right)$.
Then we can compute derivatives of $f$:
\begin{align*}
f(x) &= \arctan \left( \frac{\tan x}{2} \right)
& f(0) &= 0 \\
f'(x) &= \frac{4}{3 \cos(2x) + 5}
& f'(0) &= \frac{1}{2} \\
f''(x) &= \frac{24 \sin(2x)}{(3 \cos(2x) + 5)^2}
& f''(0) &= 0 \\
\end{align*}
If you continue in this way, you get the Taylor series for $f$:
$$
f(x) = \frac12 x + \frac18 x^3 + \frac{1}{32} x^5 + \frac{11}{1920} x^7 + \cdots
$$
In particular, for small values of $x$, the second taylor approximation
$\frac12 x$ is pretty accurate.
More precisely, if the third derivative is bounded by $M$, by the Lagrange remainder theorem, we have
$$
f(x) \approx \frac12 x
$$
with error no more than
$M \frac{x^3}{3!}$ -- an incredibly insignificant value when $x$ is small.
That is, we can approximate
$$
\theta_1 = \arctan \left( \frac{\tan \theta}{2} \right)
\approx \frac{\theta}{2}
$$
with error on the order of $\theta^3$.
