Download Definition of the Inverse Secant Function

Definition of the Inverse Secant Function
From trigonometry, we know that the secant function is defined by
1 .
secx 
cosx
Since cosx  0 at x  /2, 3/2, 5/2. etc. (all odd multiples of /2), the secant
function is not defined at these values of x (since division by zero is undefined). Also,
since the range of the cosine function is −1, 1, the range of the secant function is
−, −1  1, . A portion of the graph of the secant function is shown below. The
graph has vertical asymptotes at all odd multiples of /2.
Graph of y  secx
If we restrict the domain of the secant function to 0, /2  /2,  (as shown in the
graph below), then we have a one–to–one function which thus has an inverse.
y  secx with domain 0, /2  /2, 
The inverse secant function (denoted by arcsec) is defined using the restricted
domain on the secant function described above. Thus the inverse secant function has
domain −, −1  1,  and range 0, /2  /2, . For each x in the set
−, −1  1,  we define
y  arcsecx
to be the unique value of y in the set 0, /2  /2,  for which
secy  x.
A graph of the inverse secant function is shown below.
To make sure that we understand the definition of the arcsec function, let us consider
a couple of examples.
Example
What is the value of arcsec1?
To answer this we must realize that
y  arcsec1
means that
secy  1
and that y is a number in 0, /2  /2, . Based on our knowledge of trigonometric
functions we know that cos0  1 and hence
1
sec0 
 1  1.
1
cos0
Furthermore we see that 0 is a number in the set 0, /2  /2, . Therefore
arcsec1  0.
Why isn’t it true that arcsec1  2? After all it is true that
1
sec2 
 1  1.
1
cos2
However note that the number 2 is not in the set 0, /2  /2, . Thus
arcsec1 ≠ 2 even though sec2  1.
Example
What is the value of arcsec − 2 ?
To answer this we must realize that
y  arcsec − 2
means that
secy  − 2
and that y is a number in the set 0, /2  /2, . Based on our knowledge of
trigonometric functions we know that cos3/4  − 2 /2 and hence
1
sec3/4 
 1  − 2.
2
cos3/4
− 2
Furthermore we see that 3/4 is a number in the set 0, /2  /2, . Therefore
arcsec − 2  3/4.
Why isn’t it true that arcsec − 2  −3/4? After all it is true that
sec − 3
4

1
 1
3
2
cos− 4 
− 2
 − 2.
However note that the number −3/4 is not in the set 0, /2  /2, . Thus
arcsec − 2 ≠ −3/4 even though sec−3/4  − 2 . In fact, notice that the value of
arcsecx is never negative for any value of x. (Look at the graph of the arcsec function
shown above.)
Exercise
Determine the values of the following. (Each can be done without a
calculator – just using knowledge of the unit circle and the definition of arcsec).
1. arcsec 2 3 /3
2. arcsec
2
3. arcsec2
4. arcsec −2 3 /3
5. arcsec−2
6. arcsec−1
The Derivative of arcsec
We now compute the derivative of y  arcsecx. Rewriting y  arcsecx as secy  x
and using implicit differentiation gives
d secy  d x
dx
dx
or
dy
1
secy tany
dx
or
dy
1

.
dx
secy tany
Using the trigonometric identity
sec 2 y  1  tan 2 y
we obtain
tany   sec 2 y − 1 .
If x is in the interval 1, , then y is in the interval 0, /2 (see the graph of the arcsec
function) meaning that tany  0 and we choose the “” sign on the above square
root. This gives us
dy
1

.
2
dx
x x −1
However, if x is in the interval −, −1, then y is in the interval /2,  meaning that
tany  0 and we choose the “−” sign on the square root and this gives us
dy
1

.
dx
−x x 2 − 1
At x  1 and x  −1, the arcsec function is not differentiable. Its graph has a vertical
(infinite slope) tangent line at these points.
In conclusion, the derivative of the inverse secant function is
1
d arcsecx 
dx
−x x 2 −1
1
x x 2 −1
if −  x  −1
if
.
1x
Integrals Involving arcsec
From the above differentiation formula, we can see that the value of
 21 dx
x x −1
depends on what interval we are working with. If we are working on some interval that
is a subset of 1, , then
 21 dx  arcsecx  C.
x x −1
However if we are working on some interval that is a subset of −, −1, then
 21 dx  − arcsecx  C
x x −1
because on such an interval we have
d − arcsecx  − d arcsecx  −
dx
dx
For example, the interval

2
1
2
x x −1
2
− 2
1
x x −1
2
1

x x −1
2
.
2 , 2 is a subset of 1,  and thus
dx  arcsecx| 2 2  arcsec2 − arcsec
However, the interval −2, − 2
 −2
1
−x x 2 − 1
2
  .
12
is a subset of −, −1 and hence
dx  − arcsecx| −−2 2  − arcsec − 2
− − arcsec−2  −  .
12