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?
 A: According to the small angle approximation the tangent of an angle equals approximately the angle. 
Tangent of the old angle ABC is the quotient between AC and AB. Denote it x.
Tangent of the new angle ABC is the quotient between AC and twice AB. Denote it y.
 y = 2x
Both triangles have the angle ABC very small. So x is equal to the old ABC angle and y becomes the new ABC angle. 
So, the old angle ABC is half the new one.
A: 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 $\theta^3$.
