I'm trying to solve the following problem from Smart's Text-Book on Spherical Astronomy (exercise 5 on p.23 of the 6th ed.):
$A$ and $B$ are two places on the earth's surface with the same latitude $\phi$; the difference of longitude between $A$ and $B$ is $2l$. Prove that
- the highest latitude reached by the great circle $AB$ is $\tan^{-1} ( \tan \phi \sec l )$, and
- the distance measured along the parallel of latitude between $A$ and $B$ exceeds the great circle distance $AB$ by $$2 \csc 1' [ l \cos \phi – \sin^{-1} ( \sin l \cos \phi ) ]\text{ nautical miles}.$$
I have tried everything I can think of to solve the first part, and I fell like I'm missing something obvious. I try to solve the triangle, use trigonometric identities, sine law, polar triangles, and nothing works.
Best Answer
I think it is a little more straightforward if you work in Cartesian coordinates.
Take an earth of radius one, for simplicity.
Then we can take the longitudes as $\pm l$ without loss of generality. Then the halfway point between the two points, projected to the surface of the earth, will have the highest latitude (equivalently the highest $z$ component).
The two points are $(\cos \phi \cos l, \cos \phi \sin l, \sin \phi)$ and $(\cos \phi \cos l, -\cos \phi \sin l, \sin \phi)$, and the mid point is $(\cos \phi \cos l, 0, \sin \phi)$. To project to the surface, we divide by the norm, to get a $z$ component (on our unit earth) of $\sin \delta = {\sin \phi \over \sqrt{ (\cos \phi \cos l)^2 + \sin^2 \phi}} = { \tan \phi \over \sqrt{\cos^2l+\tan^2 \phi} }$.
To obtain the $\arctan$, we first need $\cos \delta$, which we get from $\cos \delta = \sqrt{1 -\sin^2 \delta} = { \cos l \over \sqrt{ \cos^2l+\tan^2 \phi} } $, and hence $\tan \delta = {\tan \phi \over \cos l}$, which is the desired result.