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Trigonometry, Logs and Map Projections

Date: 10/01/97 at 10:35:54
From: Gavin Lotter
Subject: Trigonometry, logs and map projections

Dear Doc,

I have a headache from trying to figure out the inverse function used 
by the Mercator map projection, a very simple map projection.

To project a point of longitude and latitude onto a piece of paper, 
the Mercator projection uses X as the longitude and Y is determined by 
the formula

        Y = log_e( sec(A) + tan(A) )

where A is the latitude.

My problem is that I need to determine the latitude from the projected 
point, that is Y is known.  The formula then becomes

        sec(A) + tan(A) = e^Y

where A is the latitude which I need to determine and Y is the 
projected latitude.

It appears as though I have two unknowns ( sec(A) and tan(A) ) with 
one equation. How can I possibly solve for A, the original unprojected 

Thanks for a great service and for one of the best sites on the Web.

Gavin Lotter

Date: 10/01/97 at 12:30:16
From: Doctor Rob
Subject: Re: Trigonometry, logs and map projections

Flattery will get you everywhere!  :-)  Thanks for the kind comment.

You need one more fact to realize that you don't really have two 
unknowns here. That is, for every angle A, 

   [sec(A)]^2 = 1 + [tan(A)]^2.  

You could use this to rewrite your equation as

   Sqrt[1 + tan(A)^2] + tan(A) = e^Y.

If you transpose tan(A) to the other side, then square, you will see 

   1 + tan(A)^2 = e^(2*y) - 2*tan(A)*e^Y + tan(A)^2.

Now subtract tan(A)^2 from both sides, and solve for tan(A):

   tan(A) = [e^(2*Y)-1]/2*e^Y,
          = [e^Y - e^(-Y)]/2,
          = sinh(Y),

(if you are familiar with hyperbolic functions).  Thus

   A = Arctan([e^Y - e^(-Y)]/2),
     = Arctan(sinh[Y]).

That ought to take care of this problem.

-Doctor Rob,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Trigonometry

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