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Expressing Arccos in Terms of Arctan

Date: 02/24/97 at 13:07:04
From: John Neubert
Subject: arccos func. in QBasic

The following is for my son. He is doing a science project involving 
asteroidal orbits for which he needs to create a Basic (QBasic) 
program to run his simulation.  He has been able to handle all the 
math and programming, but has run into a snag in that Basic only has 
arctan as a built-in function (ATN).  The books all say that from 
this, one can create the other functions.  My son needs the equation 
for arccos using arctan in its definition.  He has come up with the 
following (HOWEVER, after coding, it does not give a valid answer when 

cos x = 1 / (sec x)

(tan x)^2 + 1 = (sec x)^2 

cos x = 1 / ((tan x)^2 + 1)^.5

arccos x = 1 / ((ATN x)^2 + 1)^.5

It has been many years since I've worked with this, but his first 
three steps seem valid.  It's the last one that looses me where he 
substitutes arccos for cos and arctan (ATN) for tan.  He says this is 
valid.  Do you see any errors in the above?

When he programs it he creates a function with the QBasic "Function" 
command it produces invalid results. Can you help him?

Thank you very much.

Date: 03/07/97 at 19:21:32
From: Doctor Luis
Subject: Re: arccos func. in QBasic

Obtaining the other inverse trig functions from arctan x is not 
difficult. Drawing a right triangle with unit hypotenuse can help 
visualize the derivation:

                     / |   
                    /  |    
                   /   |      
               1  /    |       
                 /     |      
                /      |  sqrt(1-x^2)       [sqrt(z) means z^0.5]
               /       |      
              /        |   
            A           B 

Now, consider both the cosine and the tangent of angle A:

            cos A = x/1 = x
            tan A = (sqrt(1-x^2))/x

Clearly, A = arccos(x) and also A = arctan((sqrt(1-x^2))/x).

Therefore, arcos(x) = arctan((sqrt(1-x^2))/x).

The above expression gives the arccos(x) function in terms of x (which 
is given) and the arctan (or ATN) function.

I hope this helped.

-Doctor Luis,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 03/07/97 at 19:13:38
From: Doctor Ken
Subject: Re: arccos func. in QBasic

Hi -

Your instincts are correct, but the last step isn't valid.  If we 
could do this kind of thing, we could do some pretty bizarre things 
with math!  For instance, your son's step was this:

cos x = 1 / ((tan x)^2 + 1)^.5
arccos x = 1 / ((ATN x)^2 + 1)^.5

Now, if we could substitute inverse functions on both sides of an 
equation, we would be able to do things like this:

      1 = 1
  x + 1 = x + 1
  x + 1 = (x - 1) + 2   

Now plug in inverse functions - on the left, the inverse function 
of x is x, and the inverse function of x-1 is x+1:

  x + 1 = (x + 1) + 2
  x + 1 = x + 3
      1 = 3

So you can see that in general this technique won't produce valid 
results. For a geometric demonstration of how you _can_ produce a 
formula for arccos in terms of arctan, Dr. Luis' method (above) works 
just fine.

-Doctor Ken,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
Associated Topics:
High School Calculators, Computers
High School Trigonometry

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