# Trigonomic algebra to find the inverse of a relative mercator projection

Hi guys, my first post here. This isn't homework or anything, just a personal project.

## Homework Statement

I need to find the inverse of the relative mercator equation. That is, given an origin latitude/longitude, find the current latitude/longitude from an x,y point. Because longitude is easy to calculate, we will ignore that, but if anyone is interested in the formula I will post it up. I'll also post up what each variable means if anyone is interested, but I don't want to convolute this post as the context isn't important - this is just trigonimic algebra. I have tried to solve in Maple but I guess it lacks the trignomic indentity needed?

## Homework Equations

Need to rearrange the below equations for L.

x = w * log((c + s) / (c - s)) / (2 * PI) (1)
c = cos(R * (o + L) (2)
s = sin(R * (o - L) (3)

## The Attempt at a Solution

Rearranging equation 1 to find u, the "log term"
u = (c + s) / (c - s) (4)
x = w * log(u) / (2 * PI)
u = e^((2 * y * PI) / w)

Rearranging equation 2 to find L:
c = cos(R * (o + L)
L = (2 * arccos(c) / R) - o (5)

Rearranging equation 3 to find L:
s = sin(R * (o - L)
L = o - (2 * arcsin(s) / R) (6)

Rearranging equation 4 to find c:
c = ((u + 1) * s) / (u - 1) (7)

Equating 5 and 6:
(2 * arccos(c) / R) - o = o - (2 * arcsin(s) / R) (8)

Substituting equation 7 into equation 8:
(2 * arccos((u + 1) * s) / (u - 1)/ R) - o = o - (2 * arcsin(s) / R) (9)

This is where I get stuck. I need to rearrange equation 9 for s, then it will just be simple substitution in order to get the equation I need for L. I have managed to reduce it to a more readable state (you may or may not want to go down this path).

o * R = arccos((S * (U + 1)) / (U - 1)) + arcsin(S)

Now we have S stuck in an arcsin and an arccos with the latter S being involved in a more complex arc.

Any help would be very much appreciated!

Thanks,
Mickey