This geometry occurs in the definition of the angle of solar declination. Planes $X$ and $Y$ intersect along line $AF$. The angle between the two planes is defined as $a_0$ which is equivalently the angle $\angle BAC$. Now the side $AB$ is rotated to a new position denoted by $AD$ by an angle $p$ (or $\varphi$, but $p$ is what's in the image). The new angle $a_1 = \angle DAE$ is a function of $a_0$ and $p$. What is the expression for $a_1$ in terms of $a_0$ and $p$?
Note that the angle $p = \angle DAB$, all of which points lie in plane $X$. The triangles $ABC$ and $ADE$ are drawn to show the geometry but are not required to define the problem. This problem occurs between Earth and the Sun as Earth rotates around the Sun. The angle $p$ is $0$ or $180^\circ$ at solstice. The angle $a_0$ is the maximum declination angle of $23.45^\circ$.
An approximate answer appears to be $\sin(a_1) = \sin(a_0) \sin(90-p)$. How can this be proven?
@Peter gave a rigorous vector based proof. The following is a geometry based proof. Please see if the logic is correct.. The geometry is constructed by choosing vectors AC and AE with unit length. Extended line CE meets AF at point Q. Once the triangles QAB and QBC are constructed as shown, the lengths of remaining sides are established using length/sin rule for a triangle. Observe that the rotation angle "q" instead of "p" is used to determine the lengths QE and EC. Then using similarity rule for triangles the ration of sides DE:BC is equated to QE:QC. The trigonometric formula Cos(A-B)=CosACosB + SinASinB allows L1/(L1+L2) to be reduced to Cos(q).
If the rotation angle "p" is referenced instead of "q" then the proof is repeated with vectors AB and AD with unit length. In this case the lengths of sides BC and DE becomes tan(a0) and tan(a1) which will results in tan(a1)=tan(a0)*cos(p).
Does this make sense? @Peter any comment?
Since angle p and q are not identical and they are linked to each other, the results makes sense.
9/20/2013 Simplest Geometry-based Proof:
Please refer to the following figure. Objective is to link angles a0 and a1 to the rotation angle either "p" or "q". The angle between the two planes X&Y is defined as a0. Hence the plane ABC is normal to both planes X and Y. By rotating the plane ABC to a new plane ADE by amount "p" (defined by BAD in plane-X) or by amount "q" (defined by CAE in plane-Y) the new angle "a1" is obtained. Note that the plane BCED is generated by drawing it parallel to line AF while keeping it normal to plane-X. This generates the rectangle BCED. Once BCED is understood to be a rectangle the derivation of the relationship is simple as shown in the figure.
In the definition of solar declination angle, plane-X encompasses the orbit of earth around the sun and is referred to as ecliptic plane. Plane-Y is the equatorial plane which is normal to the axis of rotation of the earth. In one year as earth orbits the sun, the angle "p" evolves from 0 through 360 degrees at a constant rate of 360/365.25 degrees per day. Note that angle "p" is zero when the earth axis is tilted towards the sun on June 21 (solstice) at a maximum of 23.45 degrees (=a0). It is of interest to estimate a1 vs. p as earth orbits the sun. In the solar literature the relation ship sin(a1)=sin(a0)*cos(p) is used where as it should be tan(a1)=tan(a0)*cos(p). The angle "q" is approximately same as "p" but it will evolve not at constant rate of 360/365.25 degree/day. Is this observation correct?