Inverse Functions: Arcsec(x)

Date: 02/11/99 at 20:15:09
From: Knote
Subject: Calculus

If y = arcsec(x) then what does x equal?

Date: 02/22/99 at 15:53:11
From: Doctor Elizabeth
Subject: Re: Calculus

Hi, Knote.

Good question! Let's take a closer look at that.

First, what does the notation "arcsec()" stand for? Well, obviously it 
has something to do with trigonometry. More specifically, judging from 
the second half - "sec" - you might already know that it is somehow 
related to the sec() value of an angle. Now the question is, what is 
the relation?

Let me give you a hint: arcsec(x) is the INVERSE function of sec(x). 
Have you studied inverse functions yet? If not, I'll give you a quick 
overview. Basically, if y = f(x), then f^(-1)(x) is how we represent 
its inverse. Remember the definition of a function? It's like a 
mapping. Each value of x has a corresponding value y, after some 
operation which is represented by f. Its inverse, then, as you would 
expect, expresses the relation the other way round. For each value of 
y, we look for an x, and the operation we need to go through to get 
from y->x is the inverse of f, or f^(-1).

For example, we have the function y = 3x. When x=1, y=3. Now when y=3, 
what's value of x? x=1, right? And so how is x related to y? x=(1/3)y,  
which is the inverse function we are looking for! By convention, we 
use x as the independent variable, so we write f^(-1)(x)=(1/3)x. 

So what's special about inverse functions? If we didn't change the 
independent variable in the inverse function, what would it be? Isn't 
it x = f^(-1)(y)? And the original function is y=f(x). Now have you 
noticed something in common in these two expressions? How about a 
substitution? Let's write y as f(x) in the inverse function. Now we get 
x = f^(-1)(f(x))*.

Back to your question: arcsec(x) is the inverse of sec(x). If we plug 
them in into the equation we just got(*), we'll have x = arcsec(sec(x)) 
or x = sec(arcsec(x)), depending on whether you use sec(x) as the 
original function or the other way round. I'll leave this part for you, 
then. 

Good luck in figuring out this last step. If you still have questions, 
don't hesitate to write us back.

- Doctor Elizabeth, The Math Forum
  http://mathforum.org/dr.math/