What does it mean to multiply the precession of the earth's axis with $\sin{(23.5)}$? In an online book

Geodätische Astronomie: Grundlagen und Konzepte

by Albert Schödlbauer I saw a picture of precession and nutation of Earth's axis.

In German but I suppose it is understandable.
The precession is: $$ 360 ^\circ / (25.728  \text{ years}) = 50.2 '' \text{ per year} $$
In the book the autor states that the true rotation axis moves along a small cone wrt the average rotation axis.
This is the small elliptical cone with axes a/b in the picture.
Then he goes on and states that the true axis moves along the great precesion cone with an angular velocity of $$ 50.2'' \cdot \sin{(23.5)} = 20.1'' \text{ per year}.$$
Can someone explain why this is so, I don't get it by myself.
In another book the same is stated in the following way.   
The earth's rotation vector is inclined $23.5^\circ$ to the pole of it's orbital plane, the ecliptic. The period of the resulting precession is about $26,000$ years, corresponding to a motion of the rotation vector of $20$ arcsec per year [$2\pi \sin{(23.5^\circ) / 26,000}$ radians per year].
Why ??  
While: The line of intersection of the ecliptic and celestial equator precesses at a rate of $50$ arcsecs per year
 A: This corresponds roughly to the angle between the rotation axis in year $n$ and the rotation axis in year $n+1$. It simply has to do with the way to account for angles in spherical coordinates. This is because if you look at the circle drawn by the tip of a unit vector precessing around $z$ with a constant angle $\theta$, the radius of this circle will not be $1$ but rather $\sin(\theta)$ (this is also why the elementary solid angle in 3D is $d\theta \sin(\theta) d\phi$ and not simply $d\theta d\phi$).
If you look at the angle formed between a unit vector of angle $(\theta, \phi)$ in spherical coordinates, with a vector of angle $(\theta, \phi + d\phi)$, it will be $\sin(\theta) d\phi$. Indeed, if $\theta = \pi/2$ for instance, the vector is moving along the equator, which is a great circle on the sphere, so you can directly convert $d \phi$ into an angle. In the other extreme case, if $\theta = 0$, then no matter the value of $\phi$, the vector will be exactly vertical so that all measured angles will be $0$. In between, you get a $\sin(\theta)$ factor. 
A: So like remarked by QuantumApple it's just spherical coordinates. The angle that is meant in the referenced texts is the angle between two positions of Earth's rotation axis.
When this angle is small e.g. $\phi$ and the angle in the equatorial plane - or any other horizontal plane - is also small $\theta$, then the cosine rule gives:
$$ r^2 + r^2 - 2 \cdot r\cdot r \cos{\phi} = (r\sin{(23.5)} \cdot \theta)^2 $$
and with $\cos{\phi} = 1 - 0.5 \phi^2$ this gives $\phi = \theta \sin{(23.5)}$.
So for small angles $d\phi = 20''$, $d\theta = 50''$.
I didn't realize that during a full $360^\circ$ rotation in a horizontal plane the $23.5^\circ$ tilted axis of rotation does not turn $360^\circ$.
