= f ( x), where f is
y = f ( x)
the function (see Figure 5.3.1). There is a simple vertical rule
Figure 5.3.1
for determining whether a rule y = f ( x) is a function: f is a
function if and only if every vertical line intersects the graph
of y = f ( x) in the xy-coordinate plane at most once (see Figure 5.3.2).
y
y
y = f ( x)
y = f ( x)
x
x
(a) f is a function
(b) f is not a function
Figure 5.3.2
Vertical rule for functions
Recall that a function f is one-to-one (often written as 1 − 1) if it assigns distinct values
of y to distinct values of x. In other words, if x 1 = x 2 then f ( x 1) = f ( x 2). Equivalently, f is one-to-one if f ( x 1) = f ( x 2) implies x 1 = x 2. There is a simple horizontal rule for determining whether a function y = f ( x) is one-to-one: f is one-to-one if and only if every horizontal line
intersects the graph of y = f ( x) in the xy-coordinate plane at most once (see Figure 5.3.3).
y
y
y = f ( x)
y = f ( x)
x
x
(a) f is one-to-one
(b) f is not one-to-one
Figure 5.3.3
Horizontal rule for one-to-one functions
If a function f is one-to-one on its domain, then f has an inverse function, denoted by
f −1, such that y = f ( x) if and only if f −1( y) = x. The domain of f −1 is the range of f .
Inverse Trigonometric Functions • Section 5.3
121
The basic idea is that f −1 “undoes” what f does, and vice versa. In other words,
f −1( f ( x)) = x for all x in the domain of f , and
f ( f −1( y)) = y for all y in the range of f .
We know from their graphs that none of the trigonometric functions are one-to-one over
their entire domains. However, we can restrict those functions to subsets of their domains
where they are one-to-one. For example, y = sin x is one-to-one over the interval − π , π , as
2 2
we see in the graph below:
y
y = sin x
1
x
0
π
− π
π
− π 2
2
−1
Figure 5.3.4
y = sin x with x restricted to − π , π
2 2
For − π
we have
2 ≤ x ≤ π
2
−1 ≤ sin x ≤ 1, so we can define the inverse sine function y =
sin−1 x (sometimes called the arc sine and denoted by y = arcsin x) whose domain is the
interval [−1,1] and whose range is the interval − π , π . In other words:
2 2
sin−1(sin y) = y for − π
(5.2)
2 ≤ y ≤ π
2
sin (sin−1 x) = x for −1 ≤ x ≤ 1
(5.3)
Example 5.13
Find sin−1 sin π .
4
π
Solution: Since − π
, we know that sin−1 sin π
, by formula (5.2).
2 ≤ π
4 ≤ π
2
4
= 4
Example 5.14
Find sin−1 sin 5 π .
4
Solution: Since 5 π
, we can not use formula (5.2). But we know that sin 5 π
. Thus,
4 > π
2
4 = − 12
sin−1 sin 5 π
is, by definition, the angle y such that
and sin y
. That
4
= sin−1 − 1
− π
= − 1
2
2 ≤ y ≤ π
2
2
angle is y = − π , since
4
sin − π
.
4
= −sin π 4 = − 12
Thus, sin−1 sin 5 π
.
4
= − π 4
122
Chapter 5 • Graphing and Inverse Functions
§5.3
Example 5.14 illustrates an important point: sin−1 x should always be a number between
− π and π . If you get a number outside that range, then you made a mistake somewhere.
2
2
✄
This why in Example 1.27 in Section 1.5 we got sin−1(−0.682) = −43◦ when using the sin−1
✂
✁
button on a calculator. Instead of an angle between 0◦ and 360◦ (i.e. 0 to 2 π radians) we got
an angle between −90◦ and 90◦ (i.e. − π to π radians).
2
2
In general, the graph of an inverse function f −1 is the reflection of the graph of f around
the line y = x. The graph of y = sin−1 x is shown in Figure 5.3.5. Notice the symmetry about the line y = x with the graph of y = sin x.
y
y = sin−1 x
π
2
1
y = sin x
x
0
π
−1
1
− π 2
2
−1
y = x
− π 2
Figure 5.3.5
Graph of y = sin−1 x
The inverse cosine function y = cos−1 x (sometimes called the arc cosine and denoted
by y = arccos x) can be determined in a similar fashion. The function y = cos x is one-to-one
over the interval [0, π], as we see in the graph below:
y
y = cos x
1
x
0
π
π
− π
3 π
2
2
2
−1
Figure 5.3.6
y = cos x with x restricted to [0, π]
Thus, y = cos−1 x is a function whose domain is the interval [−1,1] and whose range is the
interval [0, π]. In other words:
Inverse Trigonometric Functions • Section 5.3
123
cos−1(cos y) = y for 0 ≤ y ≤ π
(5.4)
cos (cos−1 x) = x for −1 ≤ x ≤ 1
(5.5)
The graph of y = cos−1 x is shown below in Figure 5.3.7. Notice the symmetry about the
line y = x with the graph of y = cos x.
y
y = cos−1 x
π
1
y = cos x
x
0
π
−1
1
π
− π 2
2
y = x
−1
Figure 5.3.7
Graph of y = cos−1 x
Example 5.15
Find cos−1 cos π .
3
π
Solution: Since 0 ≤ π
, by formula (5.4).
3 ≤ π, we know that cos−1 cos π
3
= 3
Example 5.16
Find cos−1 cos 4 π .
3
Solution: Since 4 π
. Thus,
3 > π, we can not use formula (5.4).
But we know that cos 4 π
3 = − 1
2
cos−1 cos 4 π
is, by definition, the angle y such that 0
. That
3
= cos−1 − 12
≤ y ≤ π and cos y = − 12
angle is y = 2 π (i.e. 120◦). Thus, cos−1 cos 4 π
.
3
3
= 2 π
3
Examples 5.14 and 5.16 may be confusing, since they seem to violate the general rule for inverse functions that f −1( f ( x)) = x for all x in the domain of f . But that rule only applies when the function f is one-to-one over its entire domain. We had to restrict the sine and
cosine functions to very small subsets of their entire domains in order for those functions
to be one-to-one. That general rule, therefore, only holds for x in those small subsets in the
case of the inverse sine and inverse cosine.
124
Chapter 5 • Graphing and Inverse Functions
§5.3
The inverse tangent function y = tan−1 x (sometimes called the arc tangent and de-
noted by y = arctan x) can be determined similarly. The function y = tan x is one-to-one over
the interval − π , π , as we see in Figure 5.3.8:
2 2
y
3
2
1
x
0
π
π
− π 2
− π 4
4
2
y = tan x
−1
−2
−3
Figure 5.3.8
y = tan x with x restricted to − π , π
2 2
The graph of y = tan−1 x is shown below in Figure 5.3.9. Notice that the vertical asymp-
totes for y = tan x become horizontal asymptotes for y = tan−1 x. Note also the symmetry
about the line y = x with the graph of y = tan x.
y
3
y = tan x
2
π
2
1
y = tan−1 x
x
0
π
π
− π 2
− π 4
4
2
−1
− π 2
y = x
−2
−3
Figure 5.3.9
Graph of y = tan−1 x
Inverse Trigonometric Functions • Section 5.3
125
Thus, y = tan−1 x is a function whose domain is the set of all real numbers and whose
range is the interval − π , π . In other words:
2 2
tan−1(tan y) = y for − π
(5.6)
2 < y < π
2
tan (tan−1 x) = x for all real x
(5.7)
Example 5.17
Find tan−1 tan π .
4
π
Solution: Since − π
, we know that tan−1 tan π
, by formula (5.6).
2 ≤ π
4 ≤ π
2
4
= 4
Example 5.18
Find tan−1 (tan π).
Solution: Since π > π , we can not use formula (5.6). But we know that tan π
2
= 0. Thus, tan−1 (tan π) =
tan−1 0 is, by definition, the angle y such that − π
and tan y
2 ≤ y ≤ π
2
= 0. That angle is y = 0. Thus,
tan−1 (tan π) = 0 .
Example 5.19
Find the exact value of cos sin−1 − 1 .
4
Solution: Let θ = sin−1 − 1 . We know that
, so since sin θ
4
− π 2 ≤ θ ≤ π 2
= − 14 < 0, θ must be in QIV.
Hence cos θ > 0. Thus,
1 2
15
15
cos2 θ = 1 − sin2 θ = 1 − −
=
⇒
cos θ =
.
4
16
4
15
Note that we took the positive square root above since cos θ > 0. Thus, cos sin−1 − 1
.
4
=
4
Example 5.20
x
Show that tan (sin−1 x) =
for −1 < x < 1.
1 − x 2
Solution: When x = 0, the formula holds trivially, since
0
1
tan (sin−1 0) = tan 0 = 0 =
.
x
1 − 02
Now suppose that 0 < x < 1. Let θ = sin−1 x. Then θ is in QI and sin θ = x.
θ
Draw a right triangle with an angle θ such that the opposite leg has length x and
1 − x 2
the hypotenuse has length 1, as in Figure 5.3.10 (note that this is possible since
Figure 5.3.10
0 < x < 1). Then sin θ = x 1 = x. By the Pythagorean Theorem, the adjacent leg has
length
1 − x 2. Thus, tan θ =
x
.
1− x 2
If −1 < x < 0 then θ = sin−1 x is in QIV. So we can draw the same triangle except that it would be
“upside down” and we would again have tan θ =
x
, since the tangent and sine have the same sign
1− x 2
x
(negative) in QIV. Thus, tan (sin−1 x) =
for −1 < x < 1.
1 − x 2
126
Chapter 5 • Graphing and Inverse Functions
§5.3
The inverse functions for cotangent, cosecant, and secant can be determined by looking at
their graphs. For example, the function y = cot x is one-to-one in the interval (0, π), where it
has a range equal to the set of all real numbers. Thus, the inverse cotangent y = cot−1 x is
a function whose domain is the set of all real numbers and whose range is the interval (0, π).
In other words:
cot−1(cot y) = y for 0 < y < π
(5.8)
cot (cot−1 x) = x for all real x
(5.9)
The graph of y = cot−1 x is shown below in Figure 5.3.11.
y
π
π
2
y = cot−1 x
x
0
π
π
3 π
− 3 π
− π 2
− π 4
4
2
4
4
Figure 5.3.11
Graph of y = cot−1 x
Similarly, it can be shown that the inverse cosecant y = csc−1 x is a function whose
domain is | x| ≥ 1 and whose range is − π
, y
2 ≤ y ≤ π
2
= 0. Likewise, the inverse secant
y = sec−1 x is a function whose domain is | x| ≥ 1 and whose range is 0 ≤ y ≤ π, y = π .
2
π
π
csc−1(csc y) = y for − ≤ y ≤ , y = 0
(5.10)
2
2
csc (csc−1 x) = x for | x| ≥ 1
(5.11)
π
sec−1(sec y) = y for 0 ≤