Trigonometry by Michael Corral - HTML preview

= 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 ≤