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Chapter Nine
Inverse Trigonometric Functions
A problem you sometimes may have to solve is to determine how far around
the unit circle you must travel to reach a point that has a known sine, cosine
or tangent value. This is theinverse trigonometric problem and requires the
definition of new functions, the inverse trigonometric functions. There are
six of these, one for each of the trigonometric functions.
The notation used for these inverse functions can be confusing. One choice
is the f–1 notation introduced on page 9, with, for example, sin–1 denoting the
inverse sine function and cot –1 the inverse cotangent. This notation is
acceptable if you remember what the notation signifies. When you write
sin–1 x, you are denoting the inverse sine of the value x, not
, which is csc
x. Never use a power of –1 with the trigonometric functions unless you wish
to find an inverse function value. Writing the –1 power to indicate a
reciprocal would be likely to cause confusion for the reader. Instead use csc
x for the reciprocal of sin x, sec x for the reciprocal of cos x, and cot x for the
reciprocal of tan x. You can use other negative powers without
misunderstanding, like cos–2 x =
, but sec2 x is better notation.
Another way to indicate inverse trigonometric values is to use the term
arcsin x to indicate the inverse sine of x. This notation makes use of the fact
that you are looking for the length of the arc around the unit circle ending at
a point that has a sine value of x. It refers to distances around the unit circle,
but is a strange notation to use in problems that do not refer to the unit circle.
From pages 9-10, you know that in order for a function to be invertible (to
possess an inverse that is a function), the original function must assign a
unique value to every element of the domain. The graph of such an invertible
function cannot cross any horizontal line more than once. From your work
graphing all the trigonometric functions in the last chapter, you have seen that
the cyclic property possessed by each of the trigonometric functions prevents
them from being invertible.
The other inverses are given by:
arccos x = cos–1 x
arctan x = tan–1 x
arccot x = cot–1 x
arcsec x = sec–1 x
arccsc x = csc–1 x
Graph of sin x showing the failure of the
horizontal line test for invertibility.
172 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
In order to be able to define an inverse for any of the trigonometric functions,
you must make some restrictions on the domain.
For a function to be invertible it is necessary that each element of the domain
be assigned to a different value in the range. Therefore, to define an inverse
sine function, you must restrict your domain to values for which the sine
function has no repeated assignments. You also want to be sure that you can
invert all possible values of the sine function. These include all the numbers
from –1 to 1.
To be sure that your sine function has an inverse you must restrict its domain
to an interval in which every value between (and including) –1 and 1 is
assigned by the function exactly once. You can find such an interval by
looking for a piece of the graph that only increases or decreases and covers
all the values from y = –1 to y = 1. If the curve representing the graph only
increases or decreases, the function cannot return to a value a second time,
since the graph has no turns.
Note that the scale used in the graph
on the left has placed tick marks at
every multiple of
, starting at
x=
and ending at x =
. The
vertical lines (besides the y-axis) are
drawn at x =
Graph of sin x split into monotonic intervals.
Possible choices for such intervals include
decreases from 1 to –1, or
even
in which sin x steadily
in which sin x increases from –1 to 1, or
in which sin x decreases from 1 to –1. The usual choice for
sin x is to restrict the domain to
, in which sin x is increasing and for
which sin 0 = 0. Some authors use a capital initial letter in the name Arcsin x
or Sin–1 x when using this interval to indicate that this principal interval is in
use.
,
,
, and
.
CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
Over this interval sin x takes on each value between –1 and 1, so the inverse
function, sin–1 x, has a domain of [–1, 1]. The value of the inverse function
is the length of the arc (measured from (1, 0), the initial point) that ends at a
point with a second coordinate of x.
173
Note that the image of the function
becomes the domain of the inverse
function and vice-versa.
The values you should already know are those corresponding to arcs of
lengths
,
,
, and
.
Table of known values for sin x and sin–1 x.
You can obtain the graph of the function f(x) = sin
information and the usual methods.
–1
x by using the above
The domain of sin–1 x consists of the x-values between –1 and 1. The only
values that sin x can produce are between these two numbers. The inverse
sine function finds the lengths of arcs producing certain sine values. You
cannot obtain sine values outside the [–1, 1] interval.
Since 0 is a valid input value, the inverse sine function has an intercept,
(0, sin–1 0) = (0, 0). The graph passes through the origin.
You should also know the
corresponding values for the cosine
and tangent functions.
174 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
As you increase your x-values from 0 toward 1, sin–1 x increases from 0 to
, since you require greater arc lengths along the unit circle in order to
obtain larger sine values. The maximum value of the sine function is 1,
reached at an arc length of
radians.
Similarly, as you move from 0 toward –1, the values of sin–1 x become more
negative, from 0 to
, since you need distances in the negative direction
along the unit circle to obtain negative sine values.
Graph of the inverse sine function on [–1, 1].
There is no graph outside the interval [–1, 1].
The graph below shows both sin x and sin–1 x on the same axes. In this graph
you can see that the graph of sin –1 x is a reflection of the piece of the sine
function defined on the interval
.
Graph of sin–1 x on [–1, 1].
Graphs of sin x and sin–1 x on the same axes.
Graph of sin x on [–½ , ½ ].
CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
175
Now there are other arc lengths that produce the same sine values as those in
the principal interval
. A difference of
in the length of the arc
makes no difference in the value of the sine. If you reflect the entire sin x
function, your graph shows all the arc lengths that give a value of x for the
sine. However, this is not the graph of a function since it does not pass the
vertical line test. This is why you must restrict your domain. Otherwise you
would not know if the sin–1 0 was 0, ,
, or
.
Although sin
and sin
= 1, sin
= 1
= 1, the only value of
Sin–1 1 is
.
Graph of sin–1 x with unrestricted range.
For cos–1 x, you have to restrict the domain in the same way, since cos x is a
cyclic function that is not invertible over its entire domain. You again look
for an interval in which the cosine is either increasing or decreasing, and has
values that cover the entire set of values between –1 and 1.
The lines splitting the graph of cos x
into intervals where the function is
monotonic include x =
,
x =
x=
Graph of cos x split into monotonic intervals.
, the y-axis, x =
.
, and
176 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
Looking at the graph, you can see that your choices for principal interval
include
,
,
, and
.
Since the interval
includes 0, it is the one usually selected as the
principal interval even though cos x is decreasing there.
In this interval, cos–1 0 =
, so your intercept is (0,
).
Your domain is again restricted to the values from –1 to 1, since like the sine,
the cosine function can only take on these values.
To obtain positive values for cosine, you must use arc lengths that are
shorter than
, with an arc length of 0 producing the largest cosine value, 1.
For cosine values less than 0, you need arc lengths greater than
minimum cosine value of –1 reached when the arc length is
, with a
.
Since you can find the cosine at any point on the unit circle when you know
the sine (by using the identity cos 2 x = 1 – sin2 x) and because the inverse
cosine and inverse sine are related by the identity
cos–1 x =
– sin–1 x,
You can see this in the graph on the
right below, and from the identity
the inverse cosine is of less importance than the inverse sine. Note that
inverse cosine produces values between 0 and , while inverse sine values
cos x = sin (x +
are between
the cosine lags
Graph of cos–1 x.
and
.
previous chapter, which shows that
function.
Graphs of cos–1 x and sin –1 x compared
around y = ½ .
The inverse sine and inverse tangent functions are the most important of the
inverse trigonometric functions, and you will use them frequently in calculus.
) from the
behind the sine
CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
177
Checking the graph of the tangent function,
For tan x, the vertical asymptotes,
occurring at every odd multiple of
, divide the graph into intervals
over which the graph is strictly
increasing.
Graph of tan x split into monotonic intervals.
you can see that the graph is strictly increasing in the intervals
, and
,
. Note that the round parentheses bounding these
intervals indicate that the end values are not used, unlike the square brackets
in the inverse sine and inverse cosine intervals that mean the end values are
included. The interval
is usually selected for the principal
interval, since it contains the value 0.
Notice that this time the domain of the inverse tangent contains all real
numbers because the tangent can take on any positive or negative value. This
is due to the division by cos x in the definition of tangent (remember
tan x =
. When you divide by a small denominator, the quotient
becomes large, without any bound on its size.
The intercept in the principal interval is (0, tan–1 0) = (0, 0).
Large positive values of tangent are obtained by dividing positive sine values
by positive cosine values that are near zero. This happens when you use
numbers near (but less than)
in the tangent. Similarly large negative
values are obtained by dividing negative sine values by positive cosine values
near zero. This is the result of using input values just above
tangent function.
in the
178 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
Therefore, the inverse tangents of large positive numbers are near
inverse tangents of extremely negative values are near
and the
. This describes
the end behavior and yields the graph below.
Graph of tan–1 x.
The inverse secant function is rarely used, but you can sketch its graph using
the same procedure. By viewing the graph of the secant function,
Graph of sec x.
you can see that the secant never takes on any values between –1 and 1. To
obtain the graph of sec–1 x you must eliminate from the domain all x-values
between these numbers (but not –1 or 1). This leaves your domain in two
pieces,
and
.
To be sure that you have a complete set of unique secant values you invert the
secant values produced over an increasing or decreasing interval which maps
to all the possible values.
CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
Possible choices for the principal interval include
(excluding
the impossible value x =
),
(excluding x =
),
(excluding x =
), and
(excluding x =
). Notice that each of
these sets of intervals must include values on both sides of the vertical
asymptotes x =
,
,
, or
in order to have both the full set of
negative values and the full set of possible positive values. The usual choice
for the principal interval is
.
The inverse secant function has no intercept, since there is no input value for
which the secant is 0. Over the principal interval, the secant takes on the
value 1 for x = 0, and the value –1 for x = . The points (–1, ) and (1, 0)
mark ends of the graph of sec–1 x. No graph exists between these points.
To obtain secant values greater than 1 over the principal interval, you must
use input values between 0 and
. The closer your inputs get to , the larger
your secant values become. For sec–1 x, you obtain function values close to
for large positive values of x. Similarly, sec–1 x produces values between
and
for x-values less than –1, getting closer to
as the x-values become
more negative.
Graph of sec–1 x.
The inverse secant function is not used very often and neither are cot–1 x nor
csc–1 x. You can obtain their graphs the same way, first noting which
intervals you should use as your principal intervals for cot x and csc x.
179
180 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
Since sin–1 x and tan–1 x are the inverse trigonometric functions most
commonly used, the rest of the chapter will concentrate on using them, rather
than the other four.
The graphs of inverse trigonometric functions are modified in the usual ways
by stretching and translation.
Graph f(x) = 4 sin –1 2x. The first thing you must be concerned with is
the domain. Since sin –1 x is only defined for x values in the interval
[–1, 1], sin–1 2x is only defined in
. Multiplying the variable
by 2 doubles the speed with which you move through the x-values, so you
use an interval of values half as large as usual.
There is still an intercept at the origin, and the function is strictly
increasing just like sin–1 x.
Graph of 4 sin–1 2x.
Multiplying the function by 4 results in a stretching of the graph to a
height 4 times as great as before.
Graph g(x) = tan–1 (x – 3) + 2. Since the inverse tangent function has no
restrictions on its domain, g(x) does not have any either.
Replacing the x variable with x – 3 moves the graph of tan–1 x over 3 units
to the right, and adding 2 to the result moves the graph up 2 units.
The intercept of g(x) is (0, tan
increasing, just like tan–1 x.
–1
–3 + 2) = (0, .75) and the graph is
You can obtain the graphs of other inverse trigonometric functions similarly.
Graph of tan–1 (x – 3) + 2.
Graph: 3 tan–1 (x + 1) – 2.
Graph of 3 tan–1 ½(x + 1) - 2.
CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
Calculations
You can evaluate inverse trigonometric functions for input values that
produce integer multiples of
Find sin–1
,
,
,
, and
. You know that
and
.
is the sine of
radians (and also
among others). Since you are using
principal interval, your answer is
as your
. A function can only produce one
value, so this answer is the only acceptable one, even though sin
sin
, and sin
are all equal to
,
.
Find tan–1 (–1). For the tangent to have a value of 1, both the sine and
cosine must have the same values. This only happens for input values
that are multiples of
. Since the principal interval used for tan
, the only possibilities you have are
and
. At
–1
x is
the
sine and cosine have different signs, giving you a negative quotient,
which is what you want. Your answer is
Find cot (sin–1
). Here first you must find sin–1
sine function has a value of
(like
.
for
. You know that the
radians and some of its multiples
). The only value in the principal interval that gives a sine of
Find:
a. sin–1
b. tan–1
c. sec (sin–1
d. tan–1 (sin
is
. Next you calculate cot
Find sin–1 (cos
). Since cos
=
=
=
=
.
, you must find sin–1
. You
Ans.:
a.
know that the sine is equal to
at multiples of
. To obtain
, you
b.
must be in the third or fourth quadrant. Since the principal interval for the
sine function does not include the third quadrant, your value is the one in
c.
the fourth quadrant,
d.
.
)
)
181
182 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
To obtain the values of these functions for other input values use your
calculator. By using the second function for the sin , cos , and tan keys
you have access to the sin–1, cos–1, and tan–1 functions. When using sin–1 and
cos–1, be sure your input values are not outside the [–1, 1] interval of possible
values. Also be sure that your calculator is in the mode (radian or degree)
which gives you the units you want.
Find tan–1 2. Using your calculator in radian mode you obtain a value of
1.10715 radians or in degree mode 63.43°. Note that this is within the
principal interval for the tangent,
.
Find:
a. sin–1 .25
b. cos–1 .6
c. tan–1 3.5
Find sin –1 –.8. In radian mode your answer is –.927295 radians and in
degree mode, –53.13°. Your answer is within the principal interval for
the sine function.
Find cos–1 1.5. Your calculator gives you an error message indicating a
domain error. This should remind you that numbers greater than 1 are not
valid inputs into the inverse cosine function.
Applications
You can use inverse trigonometric functions to find the angles of right
triangles when you know the lengths of the sides. You can also use them to
solve equations that contain trigonometric functions.
When you know the lengths of two sides of a right triangle you can calculate
the length of the third side by using the Pythagorean Theorem. Using the
inverse trigonometric functions, you can also find the angles. When working
with right triangles, remember to put your calculator into degree mode.
If you know that the length of the hypotenuse of a right triangle is 15 and
the side opposite the angle you want (labeled x in the diagram) has a
length of 7, you can use the inverse sine function since the known sides
are the ones used to calculate the value of sin x.
sin x =
x = sin–1
=
= 27.82°
Ans.:
a. 14.48°
b. 53.13°
c. 74.05°
CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
You can find the third angle (labeled z in the diagram) either by
subtracting the sum of the two known angles (27.82° and 90°) from the
total of 180° in the triangle, or by using inverse trigonometric functions.
cos z =
=
z = cos–1
= 62.18°
In the right triangle on the right, the side opposite the unknown angle
(labeled x) has length 8 and the side adjacent to the angle x has length 5.
To find angle x, you need the inverse tangent function.
tan x =
=
x = tan–1
= 57.99°
To find sin (tan–1 x), first realize that the tangent at the unknown angle is
x=
. Therefore, you can use x for the length of the opposite side and
1 for the length of the adjacent side. By the Pythagorean Theorem, the
hypotenuse is given by
=
. Once you have the
hypotenuse, it is easy to calculate the sine using
sin (tan–1 x) =
=
To find cos (sin –1 x), let the side opposite the angle be represented by x
and the hypotenuse by 1 so that the sine value is
. The third side
(adjacent to your angle) is given by
cos (sin–1 x) =
=
. Then
=
.
183
184 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
The distance between bases on a baseball field is 90 feet and the bases
are set at right angles to each other. If a third-baseman stops a ball 20
feet behind third base, at what angle should he throw the ball in order to
beat the runner going to first base? How far would he have to throw?
Since the third baseman catches the ball 20 feet behind third base, the
distance to home is 110 feet. This is the side adjacent to his throwing
angle. The side opposite the angle is the distance from home to first base,
90 feet. To find the angle x, calculate
x = tan–1
= 39.3°.
To find the distance of the throw, z, use the Pythagorean Theorem
z=
= approx. 142 feet
or use the fact that
sin x =
sin 39.3° =
z=
= approx. 142 feet.
The following problem was used as an example in chapter 8, on page 165.
Now you can answer more questions about this problem.
Suppose you walk due north at 4 miles per hour and continue for 3 hours.
Then you turn eastward 42° and continue at 3 miles per hour for 2 more
hours before you stop. How far are you from your starting point? If you
turn around at the end of your trip and wish to head straight back to the
start, what angle must your path make to the west of due south?
CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
In the solution to the original problem in chapter 8, the lengths of the
unknown distances were y = 4.01 miles, z = 4.46 miles and x = 16.94
miles. The angle you are looking for, labeled A in the diagram, is the
same as the angle B in the large triangle. Since you know the lengths of
all three sides of this triangle, you can find B.
From the figure, sin B =
= .2367 and B = sin–1 .2367 = 13.69°.
185
186 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
Exercises
1. Find each of the following:
a. sin-1
f.
tan-1
b. tan-1
c. sec-1
d. csc-1
g. cos-1 .4
h. tan-1 10
i.
k. sin-1 sin
l.
e. cot-1 -1
cot-1 1.2
j.
cos (tan-1 –2)
sec-1 –5
m. csc (cos-1
2. Find the domains to use for the following functions:
a. f(x) = 2 sin-1 3x
b. f(x) = tan-1
x
c. f(x) = 3 cos-1 x
3. Graph each of the functions in #2.
4. Graph:
a. f(x) = 4 tan-1 x
b. f(x) = 5 + arcsin (x + 1)
c. f(x) = 7 – 3 sin-1
(x – 1)
d. f(x) = cot-1 x
e. f(x) = arcsec 3(x –
f.
)
f(x) = csc-1 2x
5. Find:
a. sin (tan-1 x)
b. sec (sin-1
)
c. tan (cos-1
)
d. csc (tan-1
)
e. cos (sin-1 x + sin-1 y)
f.
sin (2 tan-1
)
-
Use the identity on page 157.
-
Use the double-angle identity on page 157.
d.
sin-1 5(x – 2)
)
CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
187
6. Find the value of all the trigonometric functions of , where
a.
= sin-1
b.
= arctan –5
c.
= sec-1 3
d.
= arccos x
7. You have a 12-foot ladder to place against a wall in order to reach a gutter that is 8 feet off the ground. For
safety's sake, the top of the ladder should extend above the gutter. Also, you know that it is unsafe to climb
a ladder set at less than a 45° angle. How would you place the ladder in order to reach the gutter safely?
8. At what angle does a person have to start pulling on a rope in order to haul in a boat that is 6 feet lower than
the person's hands and 40 feet away?
9. The rim of a basketball goal is set 10 feet off the ground, with the rim extending back 3 feet over the court.
The foul line is placed 15 feet from the base of the goal. If a player on the foul line shoots with his hand
placed at the top of his head, find the angle that the straight line from his hand to the rim makes with the
horizontal for a 6-foot tall player. Find the angle for a 7-foot tall player.
10. Two ships left a port at the same time, one heading due north at 20 miles per hour and the other due east at
25 miles per hour. Six hours later, the northbound ship hears a distress signal from the other. The captain
decides to turn immediately to go to the assistance of the stricken ship. Through what angle should he turn
his ship in order to head directly to the other's current position? How long will it take him to reach the other
ship at 20 miles per hour.
188 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
Answers
1. a.
h. 84.29°
2. a.
3. a.
b.
c.
i.
d.
39.81°
e.
j.
101.54°
b.
f.
k.
c.
b.
d.
g. 66.42°
l.
m.
d.
c.
CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
4. a.
b.
c.
4. d.
e.
f.
5. a.
b.
c.
e.
d.
f.
6. a. sin
=
cos
=
b. sin
=
cos
=
c. sin
=
d. sin =
cos
=
tan
tan
tan
cos
=
=x
cot
= –5
=
tan
=
cot
=
sec
cot
=
=
=
csc
sec
sec
=3
cot
=
=
=
csc
csc
=
=
sec
=
csc
=
189
190 CHAPTER 9: INVERSE TRIGONOMETRIC FUNCTIONS
7. Placing the ladder so 1 foot extends beyond the gutter gives you an angle of sin-1
= 46.66°. You can get
steeper angles by extending the ladder further past the gutter.
8. The angle the rope makes with the boat is tan -1
cot-1
= 8.53°.
The angle the rope makes with the vertical is
= 81.47°.
9. For a 6-foot tall player, the rim is 4 feet higher than his hand. The angle to the front of the rim is
tan-1
= 18.43°. The rim is only 3 feet above a 7-foot tall player's hand, so his angle is tan-1
10. The ship must turn southeast at an angle of tan-1
position. The distance the ship must travel is
to travel at 20 miles per hour.
= 14.04°.
= 51.34° to put itself on a heading for the other ship's
miles, which will take about 9.6 miles