Differing Definitions of arcsec(x) Lead to Confusion over Signs

Date: 12/03/2005 at 21:36:52
From: Richard
Subject: sec(arcsec x)

Why does sec(arcsec x) equal |x|?  Similarly, why does csc(arccsc x)
equal |x|?  I was trying to prove the derivatives of arcsec and
arccsc, but when I checked the answer my book had |x| where I had x. 
I'm trying to figure out what went wrong.  Do you have any idea?   

Here is my work:

d/dx (arcsec x)
If y = arcsec x, then arcsec y = x

= 1/((sec(arcsec x))tan(arcsec x))
= 1/(x * tan y)                     sec(arcsec x)= x... right?
= 1/(x * sqrt(sec^2 y - 1))         1 + tan^2 y = sec^2 y
= 1/(x * sqrt (x^2 - 1)

Since there is |x| instead of x in the correct anwswer, I assume that 
sec(arcsec x)= |x|.  But why?

Date: 12/03/2005 at 22:57:51
From: Doctor Peterson
Subject: Re: sec(arcsec x)

Hi, Richard.

Certainly it is not true that sec(arcsec(x)) = |x|; take a specific
negative value of x and verify that the result is negative.  It's true
that there is an absolute value in the derivative, but it comes from a
different step than taking sec(arcsec(x)). 

Let me edit what you wrote to show what I think you meant, correcting
a typo and a derivative you didn't write:

  d/dx (arcsec x):
  If y = arcsec x, then sec y = x

  dy/dx = 1/(dx/dy)
  = 1/((sec(arcsec x))tan(arcsec x))
  = 1/(x * tan y)                     sec(arcsec x)= x... right?
  = 1/(x * sqrt(sec^2 y - 1))         1 + tan^2 y = sec^2 y
  = 1/(x * sqrt (x^2 -1)

The problem is not in sec(arcsec x), but in tan(arcsec x)!

While it's true that sec(arcsec x) = x for all x in the domain, the
sign of tan(y) has to be considered.  Assuming your arcsec is defined
to have a range of [0, pi], excluding pi/2, its tangent is positive in
[0, pi/2) and negative in (pi/2, pi].  You used sqrt(sec^2 y - 1),
which is always positive!  Remember that there are two square roots,
and you have to choose the appropriate one.  You want to take the
positive root when x = sec y happens to be positive, and the negative
root when x is negative, so we can write the last two lines this way,
using the "sgn" function that gives the sign of a number (+1 or -1):

  = 1/(x * sgn(x) * sqrt(sec^2 y - 1))
  = 1/(|x| * sqrt (x^2 -1)

Here, x times its own sign gives it absolute value (always positive).

Does that help?  These sign issues can be very tricky, and I can see
why this one mystified you.  The moral of the story: when you take a
square root, ALWAYS check which sign to use under which conditions.


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

Date: 12/05/2005 at 15:55:08
From: Richard
Subject: sec(arcsec x)

Hello!  Ok, I almost understand it now.  But I don't get what the
"sgn" means and why you have to multiply that by x.  Also, this whole
arcsec function has me confused.  I just realized I have never even
seen the graph of it...

Date: 12/05/2005 at 17:36:51
From: Doctor Peterson
Subject: Re: sec(arcsec x)

Hi, Richard.

I briefly defined the sgn function; here is a fuller version:

  sgn(x) = 1  if x>0
           0  if x=0
          -1  if x<0

So

  x sgn(x) = x  if x>0
             0  if x=0
            -x  if x<0

which is the same as |x|.

So what I did was to use sgn to introduce the correct sign for each 
value of x, and then merging the sgn(x) with x to get |x|.

I was just looking at an article about the unpopularity of the arcsec 
function in calculus courses, which is partly tied to the fact that 
texts differ on its exact definition--specifically what range to use 
when x < 0 (College Mathematics Journal, Nov 2005).  It is pointed out 
that if you take pi <= arcsec(x) < 3pi/2 for x <= -1, then the 
derivative is 1/[x sqrt(x^2 - 1)], but if you define it (as I do) to 
have pi/2 <= arcsec(x) <= pi for x < -1, then the derivative has the 
absolute value.

If your text asked for the derivative of arcsec(x), I hope it has 
also shown you its graph somewhere, or at least defined it as they 
are using it!  Here is some information on it:

  Inverse Secant
    http://mathworld.wolfram.com/InverseSecant.html 


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