Solar declination as a function of solar longitude

Question

I'm trying to find the relationship between solar declination and solar longitude. Solar declination is the angle between the line Earth-Sun and the equatorial plane of the Earth. Solar longitude is the angular distance traveled by the sun since the March equinox (a geocentric model makes things simpler). The sun travels in the ecliptic plane, which is tilted 23.45° relative to the equatorial plane.

Using euclidean trigonometry, I found sin δ = sin 23.45° sin λ, δ being the solar declination and λ the solar longitude. But now I realize I can't prove all my steps.

This question has been asked before (here), but the answers have not settled the issue. One of them argues that spherical trigonometry is needed. Can someone help me?

Answer 1

The geometric setting is not well defined, I hope my interpretation is right, otherwise let me know.

We have two planes, intersecting on line $r$ and forming a dihedral angle $\alpha$ ($\alpha=23.45°$). Take any point $A$ on the first plane and name $H$ its perpendicular projection on line $r$ (see figure). Another line $HD$ on the first plane makes an angle $\beta$ (the solar longitude) with $HA$ and $D$ is chosen such that $AD\parallel r$.

From $A$ and $D$ draw the perpendiculars to the first plane, intersecting the second plane at $B$ and $E$ respectively. Then $\angle AHB=\alpha$ by definition, while $\angle DHE=\delta$ must be found as a function of $\alpha$ and $\beta$.

If we take $AH=1$, then we have $$ BH={1\over\cos\alpha},\quad AB=DE=\tan\alpha,\quad AD=BE=\tan\beta,\quad HD={1\over\cos\beta}. $$

By Pythagoras' theorem we have then $$ HE=\sqrt{BH^2+BE^2}=\sqrt{{1\over\cos^2\alpha}+\tan\beta} $$ and the cosine rule applied to $HDE$ gives the desired result: $$ \cos\delta={HE^2+HD^2-DE^2\over2 HE\cdot HD}= {1\over\sqrt{1+\tan^2\alpha\cos^2\beta}}. $$