Angles projected to planes between two lines, one of which is in rolled 3D coordinate system. Let's define standard 3D cartesian coordinate system XYZ.
In the system define a line that:


*

*has a (0, 0, 0) point

*lies on the YZ plane

*has defined an angle between itself and Z axis (lets call this elevation angle, α)


Now lets copy the XYZ coordinate system (with the line) and roll it by β degrees, the revolve axis being Z axis. Lets call the rolled system X'Y'Z'.
Now when we project those two lines onto Z'X' plane, there is some angle between them. The same goes for projecting them onto Y'Z' plane. What I need to know, is how to calculate the angels between lines in those (rolled) planes.
I need pure trigonometric solution, without calculating the vectors of the lines.
 A: Note: It is advisable to make drawings while following this solution.
Suppose the line has finite length and the line joins $(0,0,0)$ to point $(0,r,h)$. Immediately, we have
$$\tan\alpha = r/h$$
We rotate the coordinate system by $\beta$ around $z$ axis. Now we have two lines: one which stays at the same location relative to the old coordinate system, another rotates together with the coordinate system.
Project now the old line onto plane $X'Y'$. Call this line $\operatorname{OPLine}[X'Y']$. Notice that $\beta$ is now the angle between this projected line and $y'$ axis. And notice also, that this project line has length $r$.
Project $\operatorname{OPLine}[X'Y']$ onto plane $X'Z'$. If we denote $a$ the length of this projected line, we write
$$a = r \sin\beta$$
Project the new line onto plane $X'Z'$. Call this line $\operatorname{NPLine}[X'Z']$. Denote $\phi_{X'Z'}$ the angle between $\operatorname{OPLine}[X'Z']$ and $\operatorname{NPLine}[X'Z']$, which is one of the things you wanted in your  question. We write
$$\tan\phi_{X'Z'} = a/h$$
Eliminating $a$, $h$ and $r$, we obtain
$$\tan\phi_{X'Z'} = \sin\beta \tan\alpha$$
The steps to obtain $\phi_{Y'Z'}$ are similar. The rest are left as exercises to the reader (as they often said).
