How to work out miles between Longitude values based on a Latitude value. We know that when Latitude is 0, the distance between Longitude values is roughly 69 miles.
When the Latitude is +/-90, Longitude values are 0 miles.
At 0 Latitude, the earths circumference is 24,902 miles.
From pole-to-pole, the earths circumference is 24,860 miles (due to the earth's ellipsoid shape).

With this information how can I work out what the distance in miles is between 2 longitude points when there latitude is equal?
 A: Firstly, let us assume that 'latitude' is found by the angle between the north pole and your position on the surface, as opposed to equal arc-lengths along the surface. No one wants to play around with elliptical integrals of the second kind on a whim.
As said by Matt B let us assume that the Earth is an ellipsoid with the semi-major axis along x and y being equal, let's call that A (which has a value of 6399592m) and the semi-minor axis along z joining the poles of length B (which has a value of  6335437m).
Let us take the trace of the ellipsoid intersecting the x-z plane and parametrise it in terms of the latitude $ \theta $. We will take the north pole to be 0° and the equator to be 90°.

For a given latitude therefore, the positions for all longitudes are found on the locus of a circle of radius $ A \sin \phi $.

As long as you measure the longitude angle in radians, the distance between two points on the same latitude is just $A \sin \phi \Delta \theta $.
If you are willing to approximate to a sphere, you can use the following result from Spherical Trig to work out the angle:
$$
\cos s = \sin^2 \phi + \cos^2 \phi \cos \Delta \theta
$$
Multiply this s by the average of the two distances above, and you have an approximate value.
If you are looking for a more empirical answer you'll need to use reference ellipsoids such as WGS84.
A: Building on Martin's post: 
Supposing earth is a perfect ellipsoid with equation: $\frac{x^2}{a^2} + \frac{y^2}{a^2} + \frac{z^2}{b^2} = 1$.
It is usually considered that you don't need a different parameter for $y$ for the earth ellipsoid, which implies that the sections by planes $z=c$ (ie constant latitude) are circles.
So, given a latitude of $\theta \in [0; \frac{\pi}{2}]$ radians, you can parametrize the ellipse in the $(Oxz)$ space by $(x,z) = (a \cos \theta, b \sin \theta)$ which gives you an equation of $z=c=b\sin \theta$.
So you're looking at the circle of equation $x^2 + y^2 = a^2 - \frac{a^2}{b^2}\times b^2 \sin ^2 \theta = a^2 \cos ^2 \theta = R ^2$.
In the end, your longitude doesn't directly depend on $b$, and for 2 points separated by a longitude of $\varphi$ at a common latitude $\theta$, you the distance on the earth surface will be $d(\varphi,\theta) = \frac{\varphi}{2\pi} \times a|\cos \theta|$
A: This can be accomplished with Thaddeus Vincenty's inverse solution or with the haversine distance formula. If simplicity and speed of calculation is more important than accuracy, then use the haversine distance formula. Otherwise, go with Vincenty's inverse solution. 

Vincenty's Inverse Solution
$\alpha$ length of the semi-major axis of the ellipsoid
$\beta$ length of the semi-minor axis of the ellipsoid
$\gamma=\frac{1}{\alpha}(\alpha-\beta)$ flattening of the ellipsoid
$x_1, x_2$ latitude of the points
in radians 
$y_1, y_2$ longitude of the points
in radians 
$\psi=y_2-y_1$ difference in longitude
$\lambda=\psi$ first and current approximation
$\lambda_0$ previous approximation
Below are some trigonometric optimizations, for $k=1,2$
\[ \tan\omega_k= (1-\gamma)\cdot\tan x_k \] 
\[ \cos\omega_k= \frac{1}{\sqrt{1+\tan^2\omega_k}} \]
\[ \sin\omega_k=\tan\omega_k\cdot\cos\omega_k \]
Now we iterate the following calculations until $\lambda-\lambda_0 > 10^{-12}$mm
\[ \sin\phi=\sqrt{(\cos\omega_2\cdot\sin\lambda)^2+(\cos\omega_1\cdot\sin\omega_2-\sin\omega_1\cdot\cos\omega_2\cdot\cos\lambda)^2} \]
\[ \cos\phi=\sin\omega_1\cdot\sin\omega_2+\cos\omega_1\cdot\cos\omega_2\cdot\cos\lambda \]
\[ \phi=\arctan\left(\frac{\sin\phi}{\cos\phi}\right) \]
\[ \sin z = \frac{\cos\omega_1\cdot\cos\omega_2\cdot\sin\lambda}{\sin\phi} \]
\[ \cos^2 z = 1-\sin^2 z \]
\[ \cos 2\phi_m = \cos\phi-\frac{2\sin\omega_1\cdot\sin\omega_2}{\cos^2 z} \]
\[ \delta = \frac{\gamma}{16}\cos^2 z\cdot(4+\gamma\cdot(4-3\cos^2 z)) \]
\[ \lambda_0 = \lambda \]
\[ \zeta=\phi+\delta\cdot\sin\phi\cdot(\cos 2\phi_m+\delta\cdot\cos\phi\cdot(-1+2\cos^2 2\phi_m)) \]
\[ \lambda = \psi+(1-\delta)\cdot\gamma\cdot\zeta\cdot\sin z \]
Once $\lambda$ converges, calculate the following
\[ \mu^2=\frac{\cos^2 z\cdot(\alpha^2-\beta^2)}{\beta^2} \]
\[ A=1+\frac{\mu^2}{16384}\left(4096+\mu^2(-768+\mu^2(320-175\mu^2))\right) \]
\[ B=\frac{\mu^2}{1024}\left(256+\mu^2(-128+\mu^2(74-47\mu^2))\right) \]
\[ C=\frac{B}{6}\cos 2\phi_m(4\sin^2\phi-3)(4\cos^2 2\phi_m-3)\]
\[ D=\cos 2\phi_m+\frac{B}{4}\left(\cos\phi(2\cos^2 2\phi_m-1)-C\right) \]
\[ \Delta\phi=B\cdot D\cdot\sin\phi \]
\[ d = A\cdot \beta\cdot(\phi-\Delta\phi) \]
Finally we now have $d$, which is the ellipsoidal distance between $(x_1, y_1)$ and $(x_2, y_2)$ in meters. To convert distance $d$ to miles, just multiply $d$ by $0.000621371$. What this algorithm lacks in speed and simplicity, is made up by its accuracy of 0.5mm!

Haversine Distance Formula
$x_1, x_2$ latitude of the points
in radians 
$y_1, y_2$ longitude of the points
in radians 
$R$ radius of the earth in meters
\[ \alpha = \sin^2\left(\frac{x_2-x_1}{2}\right)+\cos x_1\cdot\cos x_2\cdot\sin^2\left(\frac{y_2-y_1}{2}\right) \]
\[ \beta = 2\cdot{\rm atan2}(\sqrt{\alpha}, \sqrt{1-\alpha}) \]
\[ d = R\cdot\beta \] 
Now we easily have $d$, the ellipsoidal distance between $(x_1, x_2)$ and $(y_1, y_2)$ in meters. Again to convert distance $d$ to miles, just multiply $d$ by $0.000621371$. 
A: In what follows, we will use angular distance (radians). To convert from angular distance to distance in miles or kilometers, multiply the angular distance by the radius of the Earth. For most purposes, we can assume the Earth is a sphere; we will do so here. This allows us to use the formulas of spherical trigonometry.

Geodesic Distance
If you want the length of the geodesic (shortest path, not necessarily a path of constant latitude), then we can use the Spherical Law of Cosines to get
$$
\begin{align}
\cos(\text{distance})
&=\sin^2(\text{latitude})+\cos^2(\text{latitude})\cos(\Delta\text{longitude})\\
&=1-\cos^2(\text{latitude})(1-\cos(\Delta\text{longitude}))\tag{1}
\end{align}
$$
Noting that for small angles, $1-\cos(x)\approx\frac12x^2$, subtracting $(1)$ from $1$ says that
$$
\frac12\text{distance}^2\approx\cos^2(\text{latitude})\frac12\Delta\text{longitude}^2\tag{2}
$$
Therefore, we get the approximation for small distances
$$
\text{distance}\approx\cos(\text{latitude})\,\Delta\text{longitude}\tag{3}
$$

Parallel Distance
The distance along a parallel of latitude is not the shortest distance, but it is the distance traversed if one walks due east or west to the destination. This is often the distance used when asking how long a degree of longitude is.
$$
\text{distance}=\cos(\text{latitude})\Delta\text{longitude}\tag{4}
$$
Note the similarity between $(3)$ and $(4)$.
Using $1^\circ=\frac\pi{180}$ radians and a radius of $\frac{24902\text{ miles}}{2\pi}=3963.3\text{ miles}$, we get that $1^\circ$ of longitude measures
$$
69.172\text{ miles}\times\cos(\text{latitude})\tag{5}
$$
along a parallel of latitude.

Straight Line Distance
The distance along a straight line through the Earth is given by
$$
\text{distance}=2\cos(\text{latitude})\sin(\Delta\text{longitude}/2)\tag{6}
$$
Noting that for small angles, $\sin(x)\approx x$, for small distances, $(6)$ becomes $(3)$.

Parallel Distance on an Ellipsoid
Here we will not assume the Earth is a sphere, but an ellipsoid. Unless specified otherwise, latitude is assumed to be the geodetic latitude, which is dependent on the surface normal rather than the vector from the center of the Earth (geocentric latitude). The relation between geodetic latitude, $\phi$, and geocentric latitude, $\psi$, is
$$
r^2\tan(\phi)=R^2\tan(\psi)\tag{7}
$$
where $R$ and $r$ are the equatorial and polar radii. The radius of the circle formed by a parallel of latitude is
$$
\frac{R^2}{\sqrt{R^2+r^2\tan^2(\phi)}}\tag{8}
$$
Thus, the distance traversed by walking due east or due west $1^\circ$ in longitude at latitude $\phi$ is
$$
\frac\pi{180}\frac{R^2}{\sqrt{R^2+r^2\tan^2(\phi)}}\tag{9}
$$
The problem now is to compute the polar radius ($r$) from the equatorial circumference ($2\pi R$) and the polar circumference, which is given by an elliptic integral involving $R$ and $r$
$$
\begin{align}
&4\int_0^{\pi/2}\sqrt{R^2\sin^2(\theta)+r^2\cos^2(\theta)}\,\mathrm{d}\theta\\[4pt]
&=4R\int_0^{\pi/2}\textstyle{\sqrt{1-\left(1-\frac{r^2}{R^2}\right)\cos^2(\theta)}}\,\mathrm{d}\theta\\
&\sim4R\int_0^{\pi/2}\textstyle{\left[1-\frac12\left(1-\frac{r^2}{R^2}\right)\cos^2(\theta)\right]}\,\mathrm{d}\theta\\[4pt]
&=2\pi R\textstyle{\left[1-\frac14\left(1-\frac{r^2}{R^2}\right)\right]}\tag{10}
\end{align}
$$
when $r\sim R$.
Using $2\pi R=24902$ miles and $2\pi R\left[1-\frac14\left(1-\frac{r^2}{R^2}\right)\right]=24860$ miles, we get
$$
R=3963.3\text{ miles}\qquad\text{and}\qquad r=3949.9\text{ miles}\tag{11}
$$
Use $(11)$ in $(9)$ to get more precise values than $(5)$
A: This is a hint/comment with formulas.
Suppose that the Earth is a perfect sphere:
$$x^2+y^2+z^2=R^2.$$
Cutting with the plane $z=c$ and projecting on the plane $XY$:
$$x^2+y^2=R^2-c^2.$$
This circunference has radius $\sqrt{R^2-c^2}$ and lenght $2\pi\sqrt{R^2-c^2}$.
The latitude at $z=c$ can be deduced taking $y=0$:
$$x^2+c^2=R^2,$$
$$x=\sqrt{R^2-c^2}.$$
And latitude$=\arctan(\cdots)$
Generalizing to an ellipsoid should be easy.
