relationship between a great circle arc and a latitude circle arc at a given latitude My spherical geometry is a bit rusty but looking at the figure below:

... my intuition tells me that angles $\phi$ and $\theta$ (measured in radians) are connected with the following equation:
$\phi = \theta*cos(\delta)$
... where $\delta$ is the angle corresponding to the green arc.
At least the above holds true for the special cases $\delta=0$ and $\delta=\pi/2$. Does the above equation hold true in general and what is the terminology that describes the various angles and circles I've drawn?
update on terminology
It is now clear that the angles can be more properly described as follows:

*

*angle $\theta$ is the arc between points $A'$ and $B'$ on a latitude circle at latitude $\delta$ (see the drawing on this answer which makes this point clear)

*angle $\phi$ is the arc between points $A'$ and $B'$ on a great circle

 A: The distance between the endpoints is scaled down like the circle radius by a factor of $\cos\delta$. However, that rather makes
$$ \sin \frac\phi2=\cos\delta\sin \frac\theta2.$$ 
A: The accepted answer by Hagen von Eitzen is correct.
This is just to supply a proof.
The below proof uses only the basic trigonometric definitions of $cos$ and $sin$ as well as a very elementary theorem about isosceles triangles.
In the figure below points $A'$ and $B'$ have been renamed to $A$ and $B$ and the points $A$ and $B$ in the original drawing have been omitted as they are not necessary for the proof. In particular the angle $\theta$ of the original drawing (that was defined using points on the great circle) is the same as the angle $\theta$ of the new drawing that is defined using points on the latitude circle.
Refer to the figure below and observe the following elements:


*

*a great circle with center $O$ (therefore $O$ is also the center of the sphere)

*a latitude circle at latitude $\delta$ with center $O'$

*the straight line segment $AB$ (a line segment, not an arc) with point $\Gamma$ as its midpoint

*three right angles shaded grey and outlined red

*the right triangle $AO'O$ with the right angle being the one at point $O'$

*the right triangle $O'\Gamma{}A$ (which resides on the plane defined by the latitude circle) with the right angle being the one at point $\Gamma$

*the right triangle $A\Gamma{}O$ with the right angle being the one at point $\Gamma$

*angle $\phi$ is the angle $AOB$, the angle $AO\Gamma{}$ being exactly $\phi/2$

*angle $\theta$ is the angle $AO'B$, the angle $AO'\Gamma{}$ being exactly $\theta/2$
Observe that the angle $OAO'$ is identical to the angle $\delta$ and that $OA$ is a ray of the sphere, so we can write $OA=R$

We have the following equations:


*

*$O'A = R\cdot{}cos(\delta)$ ; since $OA$ is the hypotenuse and equal to $R$, and since $OAO'=\delta$ as already noted 

*$A\Gamma{}=O'A\cdot{}sin(\theta/2)$
From $1$ and $2$ we obtain:


*$A\Gamma{}=R\cdot{}cos(\delta)\cdot{}sin(\theta/2)$
We also have (from the right triangle $A\Gamma{}O$):


*$A\Gamma{}=R\cdot{}sin(\phi/2)$ ; since $OA$ is the hypotenuse and equal to $R$
From $3$ and $4$ we have:
$$sin(\frac\phi2)=cos(\delta)\cdot{}sin(\frac\theta2) $$
$\blacksquare$
A: My attempt of a proof:
In any circle, with center O, radius r, points A and B on the circumference and angle AOB = $\alpha$, the straight segment from A to B is :
$AB_{straight}^2 = 2 r^2 \times (1-cos(\alpha))$ (1)
This is a straight application of the cosine theorem.
We also know for any angle $\alpha$ that 
$cos(\alpha) = 1-2 sin^2 (\frac{\alpha}{2})$ (2)
By combining (1), (2) we get :
$AB_{straight}^2 = 4 r^2 \times sin^2(\frac{\alpha}{2})$ (3)
Apply (3) twice on the segment $A'B'$ from the original diagram by Marcus Junius Brutus :


*

*once for the great circle with $r=R$ and $\alpha = \phi$

*once for the latitude circle with $r=R \times cos(\delta)$ and $\alpha=\theta$
That gives 
$4 R^2 \times sin^2(\frac{\phi}{2})$ = $4 (R \times cos(\delta))^2 \times sin^2(\frac{\theta}{2})$ 
which simplifies to
$sin(\frac{\phi}{2}) = cos(\delta) \times sin(\frac{\theta}{2})$
