Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Title: Trignometry Full length books
Description: It is a book based on trignometry. It contains full trignometry.
Description: It is a book based on trignometry. It contains full trignometry.
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
TRIGONOMETRY
MICHAEL CORRAL
Trigonometry
Michael Corral
Schoolcraft College
About the author:
Michael Corral is an Adjunct Faculty member of the Department of Mathematics at
Schoolcraft College
...
A
...
A
...
S
...
A
This text was typeset in L TEX with the KOMA-Script bundle, using the GNU Emacs
text editor on a Fedora Linux system
...
Copyright © 2009 Michael Corral
...
3 or any later version published by the Free
Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover
Texts
...
”
Preface
This book covers elementary trigonometry
...
The prerequisites are high school
algebra and geometry
...
There are exercises at the end
of each section
...
An average student should be able to do most of the exercises
...
This text probably has a more geometric feel to it than most current trigonometry texts
...
I think that approaching the
subject with too much of an analytic emphasis is a bit confusing to students
...
This book starts with the “old-fashioned” right triangle
approach to the trigonometric functions, which is more intuitive for students to grasp
...
So this book presents material in a very different order than most books
today
...
That seems like a more natural progression of
topics, instead of leaving general triangles until the end as is usually the case
...
Instead of taking the (doomed) approach that
students have to be shown that trigonometry is “relevant to their everyday lives” (which
inevitably comes off as artificial), this book has a different mindset: preparing students
to use trigonometry as it is used in other courses
...
Students are far more likely to need trigonometry in other courses
(e
...
engineering, physics)
...
In Chapter 5 students are asked to use the free open-source software Gnuplot to graph
some functions
...
Appendix B contains a brief tutorial on Gnuplot
...
There are a few code samples in Chapter 6, written in the Java and Python programming languages, hopefully sufficiently clear
so that the reader can figure out what is being done even without knowing those languages
...
This book probably discusses numerical issues more
than most texts at this level (e
...
the numerical instability of Heron’s formula for the area
of a triangle, the secant method for solving trigonometric equations)
...
I wanted to keep this book as brief as possible
...
However, some decisions had to be made on what material to leave out
...
The hardest decision was to exclude Paul Rider’s clever geometric
proof of the Law of Tangents without using any sum-to-product identities, though I do give
a reference to it
...
For more details, see
the included copy of the GFDL
...
The PDF version will always be freely available to the public at no cost (go
to http://www
...
net/trig)
...
edu for
any questions on this or any other matter involving the book (e
...
comments, suggestions,
corrections, etc)
...
July 2009
Livonia, Michigan
M ICHAEL C ORRAL
Contents
Preface
iii
1
1
Right Triangle Trigonometry
1
...
2
1
...
4
1
...
The Law of Cosines
...
The Area of a Triangle
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
1
...
14
...
32
38
...
...
...
...
...
Basic Trigonometric Identities
...
Double-Angle and Half-Angle Formulas
Other Identities
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
Arc Length
...
Circular Motion: Linear and Angular Speed
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
65
71
78
82
87
87
90
95
100
103
Graphing the Trigonometric Functions
...
109
Inverse Trigonometric Functions
...
1
6
...
...
...
1
5
...
3
6
...
...
...
1
4
...
3
4
...
...
...
Identities
3
...
2
3
...
4
4
...
...
...
1
2
...
3
2
...
5
3
Angles
...
Trigonometric Functions of Any Angle
...
129
Solving Trigonometric Equations
...
133
v
vi
C ONTENTS
6
...
4
Complex Numbers
...
146
Appendix A:
Answers and Hints to Selected Exercises
152
Appendix B:
Graphing with Gnuplot
155
GNU Free Documentation License
160
History
168
Index
169
1 Right Triangle Trigonometry
Trigonometry is the study of the relations between the sides and angles of triangles
...
Though the ancient Greeks, such as Hipparchus
and Ptolemy, used trigonometry in their study of astronomy between roughly 150 B
...
- A
...
200, its history is much older
...
C
...
Before discussing those
functions, we will review some basic terminology about angles
...
1 Angles
Recall the following definitions from elementary geometry:
(a) An angle is acute if it is between 0◦ and 90◦
...
(c) An angle is obtuse if it is between 90◦ and 180◦
...
(a) acute angle
(b) right angle
Figure 1
...
1
(c) obtuse angle
(d) straight angle
Types of angles
In elementary geometry, angles are always considered to be positive and not larger than
360◦
...
2 The following definitions will be used
throughout the text:
1 Ahmes claimed that he copied the papyrus from a work that may date as far back as 3000 B
...
2 Later in the text we will discuss negative angles and angles larger than 360◦
...
1
(a) Two acute angles are complementary if their sum equals 90◦
...
(b) Two angles between 0◦ and 180◦ are supplementary if their sum equals 180◦
...
(c) Two angles between 0◦ and 360◦ are conjugate (or explementary) if their sum equals
360◦
...
∠B
∠B
∠A
∠A
(a) complementary
∠A
∠B
(b) supplementary
(c) conjugate
Figure 1
...
2 Types of pairs of angles
Instead of using the angle notation ∠ A to denote an angle, we will sometimes use just a
capital letter by itself (e
...
A , B, C ) or a lowercase variable name (e
...
x, y, t)
...
1
The Greek alphabet
Letters
Name
Letters
Name
Letters
Name
A
B
Γ
∆
E
Z
H
Θ
alpha
beta
gamma
delta
epsilon
zeta
eta
theta
I
K
Λ
M
N
Ξ
O
Π
iota
kappa
lambda
mu
nu
xi
omicron
pi
P
Σ
T
Υ
Φ
X
Ψ
Ω
rho
sigma
tau
upsilon
phi
chi
psi
omega
α
β
γ
δ
ǫ
ζ
η
θ
ι
κ
λ
µ
ν
ξ
o
π
ρ
σ
τ
υ
φ
χ
ψ
ω
In elementary geometry you learned that the sum of the angles in a triangle equals 180◦ ,
and that an isosceles triangle is a triangle with two sides of equal length
...
Thus, in a right triangle one of the angles
is 90◦ and the other two angles are acute angles whose sum is 90◦ (i
...
the other two angles
are complementary angles)
...
1
3
Example 1
...
For example, in the
first triangle above we will simply refer to the angle ∠ BAC as angle A
...
⇒
For the right triangle △ DEF , E = 53◦ and F = 90◦ , and we know that the two acute angles D and E
are complementary, so
D + E = 90◦
D = 90◦ − 53◦
⇒
⇒
D = 37◦
...
2
Thales’ Theorem states that if A , B, and C are (distinct) points on a circle such that the line segment
AB is a diameter of the circle, then the angle ∠ ACB is a right angle (see Figure 1
...
3(a))
...
C
C
αβ
A
O
(a)
Figure 1
...
3
B
A
α
β
O
B
(b)
Thales’ Theorem: ∠ ACB = 90◦
To prove this, let O be the center of the circle and draw the line segment OC , as in Figure 1
...
3(b)
...
Since AB is a diameter of the circle, O A and OC have the same
length (namely, the circle’s radius)
...
Likewise, △ OBC is an isosceles triangle and ∠ OCB = ∠ OBC = β
...
And since the angles of △ ABC must add up to 180◦ , we see that 180◦ = α + (α + β) + β =
2 (α + β), so α + β = 90◦
...
QED
4
Chapter 1 • Right Triangle Trigonometry
§1
...
For example, in
c
Figure 1
...
4 the right angle is C , the hypotenuse is the line segment
a
AB, which has length c, and BC and AC are the legs, with lengths a
and b, respectively
...
Figure 1
...
4
By knowing the lengths of two sides of a right triangle, the length of
the third side can be determined by using the Pythagorean Theorem:
Theorem 1
...
Pythagorean Theorem: The square of the length of the hypotenuse of a
right triangle is equal to the sum of the squares of the lengths of its legs
...
1
...
1)
Let us prove this
...
1
...
B
c
C
d
D
c−
A
B
a
d
b
C
(a) △ ABC
a
b
d
D
C
A
(b) △ CBD
Figure 1
...
5
D
c−d
(c) △ ACD
Similar triangles △ ABC , △ CBD , △ ACD
Recall that triangles are similar if their corresponding angles are equal, and that similarity
implies that corresponding sides are proportional
...
Since △ ABC is similar to △ ACD , comparing horizontal legs and hypotenuses gives
c
b
=
c−d
b
⇒
b2 = c2 − cd = c2 − a2
⇒
a2 + b 2 = c 2
...
For example, in the above proof we had CD ⊥ AB and △ ABC ∼ △ CBD ∼ △ ACD
...
1
5
Example 1
...
For triangle △ DEF , the Pythagorean Theorem says that
e2 + 12 = 22
e2 = 4 − 1 = 3
⇒
e =
⇒
3
...
Example 1
...
At
what height is the top of the ladder touching the wall?
Solution: Let h be the height at which the ladder touches the wall
...
Then we see that the ladder, ground, and wall form a right triangle with a
hypotenuse of length 17 ft (the length of the ladder) and legs with lengths 8 ft and
h ft
...
Exercises
For Exercises 1-4, find the numeric value of the indicated angle(s) for the triangle △ ABC
...
Find B if A = 15◦ and C = 50◦
...
Find C if A = 110◦ and B = 31◦
...
Find A and B if C = 24◦ , A = α, and B = 2α
...
Find A , B, and C if A = β and B = C = 4β
...
5
...
6
...
7
...
8
...
9
...
What is the straight distance between the
car’s starting point and end point?
6
Chapter 1 • Right Triangle Trigonometry
§1
...
One end of a rope is attached to the top of a pole 100 ft high
...
11
...
(Hint: Is a2 + b2 > a2 ?)
12
...
13
...
The triple is normally written as (a, b, c)
...
(a) Show that (6,8,10) is a Pythagorean triple
...
How
would you interpret this geometrically?
(c) Show that (2 mn, m2 − n2 , m2 + n2 ) is a Pythagorean triple for all integers m > n > 0
...
Write
down the first ten Pythagorean triples generated by this formula, i
...
use: m = 2 and n = 1;
m = 3 and n = 1, 2; m = 4 and n = 1, 2, 3; m = 5 and n = 1, 2, 3, 4
...
This exercise will describe how to draw a line through any point outside a circle such that the
line intersects the circle at only one point
...
1
...
tangent line
A
•
ge
tan
not
nt
O
P
C
Figure 1
...
6
On a sheet of paper draw a circle of radius 1 inch, and call the center of that circle O
...
5 inches away from O
...
1
...
Let A be one of the points where this circle intersects the first
circle
...
In general the tangent line through a point on a circle
is perpendicular to the line joining that point to the center of the circle (why?)
...
Does it match the physical measurement of P A ?
15
...
Now imagine that you rotate the circle 180◦
around its center, so that △ ABC is in a new position, as indicated by
the dashed lines in the picture
...
C
A
O
B
Trigonometric Functions of an Acute Angle • Section 1
...
2 Trigonometric Functions of an Acute Angle
Table 1
...
For the acute
angle A , call the leg BC its opposite side, and call the leg AC its
adjacent side
...
The ratios of sides of a right triangle occur often enough in practical applications to warrant their own names, so we define the six
trigonometric functions of A as follows:
C
The six trigonometric functions of A
Abbreviation
Definition
sine A
sin A
=
opposite side
hypotenuse
=
a
c
cosine A
cos A
=
adjacent side
hypotenuse
=
b
c
tangent A
tan A
=
opposite side
adjacent side
=
a
b
cosecant A
csc A
=
hypotenuse
opposite side
=
c
a
secant A
sec A
=
hypotenuse
adjacent side
=
c
b
cotangent A
cot A
=
adjacent side
opposite side
=
b
a
We will usually use the abbreviated names of the functions
...
2 that
the pairs sin A and csc A , cos A and sec A , and tan A and cot A are reciprocals:
csc A =
1
sin A
sec A =
1
cos A
cot A =
1
tan A
sin A =
1
csc A
cos A =
1
sec A
tan A =
1
cot A
8
Chapter 1 • Right Triangle Trigonometry
§1
...
5
For the right triangle △ ABC shown on the right, find the values of all six trigonometric functions of the acute angles A and B
...
For angle A , the opposite side
BC has length 3 and the adjacent side AC has length 4
...
Thus:
sin B =
opposite
4
=
hypotenuse
5
cos B =
adjacent
3
=
hypotenuse
5
tan B =
opposite
4
=
adjacent
3
csc B =
5
hypotenuse
=
opposite
4
sec B =
hypotenuse
5
=
adjacent
3
cot B =
adjacent
3
=
opposite
4
Notice in Example 1
...
This raises the
possibility that our answers depended on a triangle of a specific physical size
...
The American student thinks that the lengths 3, 4, and
5 in Example 1
...
Since 1 in ≈ 2
...
2
...
B
B
5
B′
3
5
A
4
(a) Inches
C
A
′
3
4
C
B′
A
′
A′
(b) Centimeters
Figure 1
...
1
C′
C
(c) Similar triangles
△ ABC ∼ △ A ′ B′ C ′
If the American triangle is △ ABC and the German triangle is △ A ′ B′ C ′ , then we see
from Figure 1
...
1 that △ ABC is similar to △ A ′ B′ C ′ , and hence the corresponding angles
Trigonometric Functions of an Acute Angle • Section 1
...
In fact, we know that common ratio: the sides of △ ABC are approximately 2
...
So when the American student calculates sin A and the German student
calculates sin A ′ , they get the same answer:3
△ ABC ∼ △ A ′ B′ C ′
⇒
BC
AB
= ′ ′
′ C′
B
AB
⇒
BC
B′ C ′
= ′ ′
AB
AB
⇒
sin A = sin A ′
Likewise, the other values of the trigonometric functions of A and A ′ are the same
...
This leads us to
the following conclusion:
When calculating the trigonometric functions of an acute angle A , you may
use any right triangle which has A as one of the angles
...
This means
that the values of the trigonometric functions are unitless numbers
...
5, that is the same as the 3/5 that
the German student calculated, despite the different units for the lengths of the sides
...
6
Find the values of all six trigonometric functions of 45◦
...
Since the two legs
of the triangle △ ABC have the same length, △ ABC is an isosceles triangle,
which means that the angles A and B are equal
...
By the Pythagorean Theorem, the
length c of the hypotenuse is given by
c2 = 12 + 12 = 2
⇒
1
2
B
1
45◦
A
1
C
2
...
For example, if we multiply each side of △ ABC by 2, then we would have a similar
triangle with legs of length 2 and hypotenuse of length 2
...
The same goes for the other functions
...
10
Chapter 1 • Right Triangle Trigonometry
§1
...
7
Find the values of all six trigonometric functions of 60◦
...
Since the original triangle
3
was an equilateral triangle (i
...
all three sides had the same length), its
◦
three angles were all the same, namely 60
...
So
2
as in the figure on the right, the triangle △ ABC has angle A = 60◦ and
angle B = 30◦ , which forces the angle C to be 90◦
...
We see that the hypotenuse has length c = AB = 2 and the leg AC has length b = AC = 1
...
a =
Thus, using the angle A we get:
sin 60◦ =
csc 60◦ =
opposite
3
=
hypotenuse
2
2
hypotenuse
=
opposite
3
adjacent
1
=
hypotenuse
2
cos 60◦ =
sec 60◦ =
tan 60◦ =
opposite
3
=
=
adjacent
1
hypotenuse
=2
adjacent
cot 60◦ =
adjacent
1
=
opposite
3
3
Notice that, as a bonus, we get the values of all six trigonometric functions of 30◦ , by using angle
B = 30◦ in the same triangle △ ABC above:
sin 30◦ =
1
opposite
=
hypotenuse
2
csc 30◦ =
hypotenuse
=2
opposite
cos 30◦ =
sec 30◦ =
adjacent
3
=
hypotenuse
2
tan 30◦ =
2
hypotenuse
=
adjacent
3
cot 30◦ =
opposite
1
=
adjacent
3
3
adjacent
=
=
opposite
1
3
Example 1
...
Find the values of the other trigonometric
3
functions of A
...
The reason is that the trigonometric functions were defined in terms of
ratios of sides of a right triangle, and you are given one such function (the sine,
in this case) already in terms of a ratio: sin A = 2
...
The adjacent side to A has unknown
3
length b, but we can use the Pythagorean Theorem to find it:
22 + b2 = 32
⇒
b2 = 9 − 4 = 5
⇒
b =
5
Trigonometric Functions of an Acute Angle • Section 1
...
5 and 1
...
Generalizing those examples gives us the following theorem:
Theorem 1
...
Cofunction Theorem: If A and B are the complementary acute angles in a
right triangle △ ABC , then the following relations hold:
sin A = cos B
sec A = csc B
tan A = cot B
sin B = cos A
sec B = csc A
tan B = cot A
We say that the pairs of functions { sin, cos }, { sec, csc }, and { tan, cot } are cofunctions
...
That is how the functions cosine, cosecant, and cotangent got the
“co” in their names
...
Example 1
...
Solution: (a) The complement of 65◦ is 90◦ − 65◦ = 25◦ and the cofunction of sin is cos, so by the
Cofunction Theorem we know that sin 65◦ = cos 25◦
...
(c) The complement of 59◦ is 90◦ − 59◦ = 31◦ and the cofunction of tan is cot, so tan 59◦ = cot 31◦
...
2
...
We can use the Pythagorean
Theorem to generalize the right triangles in Examples 1
...
7 and see what any 45 −
45 − 90 and 30 − 60 − 90 right triangles look like, as in Figure 1
...
2 above
...
2
Example 1
...
D
Solution: Since 75◦ = 45◦ +30◦ , place a 30−60−90 right triangle
△ ADB with legs of length 3 and 1 on top of the hypotenuse
of a 45 − 45 − 90 right triangle △ ABC whose hypotenuse has
length 3, as in the figure on the right
...
2
...
So AC = BC =
hypotenuse divided by
3
2
3
2
...
Since
∠ BAC = 45◦ and ∠ D AB = 30◦ , we see that ∠ D AE = 75◦ since
it is the sum of those two angles
...
Notice that ∠ ADE = 15◦ , since it is the complement of ∠ D AE
...
Draw
BF perpendicular to DE , so that △ DFB is a right triangle
...
Also, ∠ DBF = 45◦ since it is the complement of
∠ BDF
...
B
F
2
3
3
2
◦
30
A
45◦
E
C
3
2
2
Now, we know that DE ⊥ AC and BC ⊥ AC , so FE and BC are parallel
...
Thus, FBCE is a rectangle, since ∠ BCE
is a right angle
...
and FE = BC =
3+1
2
+
3
2
6+ 2
4
, cos 75◦ =
=
Note: Taking reciprocals, we get csc 75◦ =
and
AE
AD
=
4
,
6+ 2
Hence,
−
1
2
, and tan 75◦ =
DE
AE
AE = AC − EC =
3−1
2
2
=
6− 2
4
sec 75◦ =
4
,
6− 2
3
2
=
=
and cot 75◦ =
3−1
2
3+1
2
3−1
2
...
6− 2
...
2
...
1
...
a = 8, b = 15, c = 17
3
...
a = 20, b = 21, c = 29
5
...
a = 1, b = 2, c =
8
...
a = 5, c = 6
5
7
...
2
...
b = 7, c = 8
For Exercises 11-18, find the values of the other five trigonometric functions of the acute angle A
given the indicated value of one of the functions
...
2
11
...
cos A =
2
3
13
...
tan A =
5
9
16
...
sec A =
2
10
14
...
csc A = 3
For Exercises 19-23, write the given number as a trigonometric function of an acute angle less than
45◦
...
sin 87◦
20
...
cos 46◦
22
...
sec 77◦
For Exercises 24-28, write the given number as a trigonometric function of an acute angle greater
than 45◦
...
sin 1◦
25
...
tan 26◦
27
...
csc 43◦
29
...
7 we found the values of all six trigonometric functions of 60◦ and 30◦
...
For an acute angle A , can sin A be larger than 1? Explain your answer
...
For an acute angle A , can cos A be larger than 1? Explain your answer
...
For an acute angle A , can sin A be larger than tan A ? Explain your answer
...
If A and B are acute angles and A < B, explain why sin A < sin B
...
If A and B are acute angles and A < B, explain why cos A > cos B
...
Prove the Cofunction Theorem (Theorem 1
...
(Hint: Draw a right triangle and label the angles
and sides
...
Use Example 1
...
A
37
...
2
...
D
(a) Find sin A
...
)
3
(b) Find the length of AC
...
(d) Figure 1
...
4 is drawn to scale
...
Is the calculator result close to
your answer from part(a)?
Note: Make sure that your calculator is in degree mode
...
2
...
In Exercise 37, verify that the area of △ ABC equals
1
2 AB · CD
...
In Exercise 37, verify that the area of △ ABC equals
1
2 AB · AC
Why does this make sense?
sin A
...
In Exercise 37, verify that the area of △ ABC equals 1 (BC )2 cot A
...
3
1
...
g
...
Today, trigonometry is widely used in physics, astronomy, engineering, navigation, surveying, and various fields of mathematics and other disciplines
...
Your calculator
should be in degree mode for these examples
...
11
A person stands 150 ft away from a flagpole and measures an angle of elevation of 32◦ from his
horizontal line of sight to the top of the flagpole
...
What is the height of the flagpole?
Solution: The picture on the right describes the situation
...
6249) = 94
...
6249 ? By using a calculator
...
Thus, the height of the flagpole is h + 6 = 94 + 6 = 100 ft
...
12
A person standing 400 ft from the base of a mountain measures the angle of elevation from the ground
to the top of the mountain to be 25◦
...
How tall is the mountain?
Solution: We will assume that the ground is flat and not inclined relative to the base of the mountain
...
Then we see that
h
20◦
h
= tan 25◦
x + 400
⇒
h = ( x + 400) tan 25 , and
h
= tan 20◦
x + 400 + 500
⇒
25◦
500
400
h = ( x + 900) tan 20◦ , so
◦
x
( x + 400) tan 25◦ = ( x + 900) tan 20◦ , since they both equal h
...
4663) = 829 ft
Applications and Solving Right Triangles • Section 1
...
13
A blimp 4280 ft above the ground measures an angle of depression of 24◦ from its horizontal line of
sight to the base of a house on the ground
...
Since the ground and the
blimp’s horizontal line of sight are parallel, we know from elementary
geometry that the angle of elevation θ from the base of the house to
the blimp is equal to the angle of depression from the blimp to the
base of the house, i
...
θ = 24◦
...
tan 24◦
Example 1
...
23◦
to the ocean horizon
...
Solution: We will assume that the earth is a sphere
...
Let the point A represent the top of the
mountain, and let H be the ocean horizon in the line of sight from
A , as in Figure 1
...
1
...
e
...
Let θ be the angle ∠ AOH
...
Also,
OH = r
...
23◦ = 87
...
We see that the line through A
and H is a tangent line to the surface of the earth (considering the
surface as the circle of radius r through H as in the picture)
...
1, AH ⊥ OH and hence ∠ OH A = 90◦
...
77◦ = 2
...
Thus,
cos θ =
OH
r
=
OA
r+3
⇒
A
B
2
...
3
...
23◦ ,
r+3
so solving for r we get
r = ( r + 3) cos 2
...
23◦ = 3 cos 2
...
23◦
1 − cos 2
...
3 miles
...
6 miles
...
The earth is an ellipsoid, i
...
egg-shaped, with an observed ellipticity
of 1/297 (a sphere has ellipticity 0)
...
26-27 in W
...
M UNK AND G
...
F M AC D ONALD, The Rotation of the
Earth: A Geophysical Discussion, London: Cambridge University Press, 1960
...
3
Example 1
...
Let O be the center of the earth, let A be a point
on the equator, and let B represent an object (e
...
a star) in space, as in the
picture on the right
...
The equatorial parallax of the sun has been observed to be approximately α = 0
...
Use this to estimate the distance from the center of
the earth to the sun
...
We want to find the length of OB
...
14, to get O A = 3956
...
Since ∠ O AB = 90◦ , we have
OA
= sin α
OB
⇒
OB =
B
α
A
O
OA
3956
...
00244◦
so the distance from the center of the earth to the sun is approximately 93 million miles
...
In the above example we used a very small angle (0
...
A degree can be divided into
smaller units: a minute is one-sixtieth of a degree, and a second is one-sixtieth of a minute
...
For example, 4
...
And
4
...
505◦
60
3600
In Example 1
...
00244◦ ≈ 8
...
Example 1
...
Use this to estimate the radius of the sun
...
The observer’s lines of sight to the visible edges of the sun
A
are tangent lines to the sun’s surface at the points A and B
...
The radius of the sun equals AS
...
So since EB = E A (why?),
1
the triangles △ E AS and △ EBS are similar
...
26722
...
In Example 1
...
Since we treated the sun in that example as a point, then we are justified in treating that distance as the distance between the centers of the earth and sun
...
6 = 92904437
...
Hence,
sin (∠ AES ) =
AS
ES
⇒
AS = ES sin 0
...
4) sin 0
...
Note: This answer is close to the sun’s actual (mean) radius of 432, 200 miles
...
3
17
You may have noticed that the solutions to the examples we have shown required at least
one right triangle
...
Often no right triangle will be
immediately evident, so you will have to create one
...
When the problem contains a circle, you can create right
angles by using the perpendicularity of the tangent line to the circle at a point5 with the line
that joins that point to the center of the circle
...
14, 1
...
16
...
17
The machine tool diagram on the right shows a symmetric V-block,
in which one circular roller sits on top of a smaller circular roller
...
Find the diameter d of the large roller, given the information in the diagram
...
3
8
Solution: The diameter d of the large roller is twice the radius
D
OB, so we need to find OB
...
C
The length OB will then be simple to determine
...
By symmetry, since the vertical line through the
centers of the rollers makes a 37◦ angle with each slanted side,
A
we have ∠ O AD = 37◦
...
So ∠ DO A = 53
...
Thus,
△ OBC is a right triangle
...
Now,
OB = OD (since they each equal the radius of the large roller), so by the Pythagorean Theorem we
have BC = DC :
BC 2 = OC 2 − OB2 = OC 2 − OD 2 = DC 2
⇒
BC = DC
Thus, △ OBC and △ ODC are congruent triangles (which we denote by △ OBC ∼ △ ODC ), since
=
their corresponding sides are equal
...
So in particular,
∠ BOC = ∠ DOC
...
Thus,
53◦ = ∠ DOB = ∠ BOC + ∠ DOC = ∠ BOC + ∠ BOC = 2 ∠ BOC
⇒
∠ BOC = 26
...
Likewise, since BP = EP and ∠ PBC = ∠ PEC = 90◦ , △ BPC and △ EPC are congruent right triangles
...
But we know that BC = DC , and we see from the diagram that EC + DC = 1
...
Thus, BC + BC = 1
...
69
...
69
= 1
...
5◦
Hence, the diameter of the large roller is d = 2 × OB = 2 (1
...
768
...
18
Chapter 1 • Right Triangle Trigonometry
§1
...
18
A slider-crank mechanism is shown in Figure 1
...
2 below
...
piston
a
A
e
conn
b
C
c
ro
cting
d
θ
r
cr
an
k
B
O
Figure 1
...
2
Slider-crank mechanism
The point A is the center of the connecting rod’s wrist pin and only moves vertically
...
As the crank rotates it makes an angle θ with the line O A
...
From Figure 1
...
2 we see
that ∠ O AC = 90◦ , and we let a = AC , b = AB, and c = BC
...
Applications and Solving Right Triangles • Section 1
...
You can think of those lengths
r cos θ
as the horizontal and vertical “components” of the hypotenuse
...
From this we determined the lengths of
the other two sides, and the other acute angle is just the complement of the known acute
angle
...
Solving a triangle
means finding the unknown parts based on the known parts
...
Example 1
...
3
...
Solving yields:
◦
a = c sin A = 10 sin 22 = 3
...
3
...
27
B = 90◦ − A = 90◦ − 22◦ = 68◦
(b) b = 8, A = 40◦
Solution: The unknown parts are a, c, and B
...
71
b
= cos A
c
⇒
c =
b
8
=
= 10
...
By the Pythagorean Theorem,
c =
a2 + b 2 =
32 + 42 =
25 = 5
...
75
...
In our case, we want the angle A such that tan A = 0
...
75
tan
Press: ✂ −1 ✁ Answer: 36
...
87◦ , approximately
...
87◦ = 53
...
✄
sin
cos
Note: The ✂ −1 ✁and ✂ −1 ✁keys work similarly for sine and cosine, respectively
...
20
Chapter 1 • Right Triangle Trigonometry
§1
...
From a position 150 ft above the ground, an observer in a building measures angles of depression of 12◦ and 34◦ to the top and
bottom, respectively, of a smaller building, as in the picture on
the right
...
34◦
12◦
150
2
...
12: A person standing a ft from the base of
a mountain measures an angle of elevation α from the ground to
the top of the mountain
...
Assuming the ground is level, find
a formula for the height h of the mountain in terms of a, b, α,
and β
...
As the angle of elevation from the top of a tower to the sun decreases from 64◦ to 49◦ during the
day, the length of the shadow of the tower increases by 92 ft along the ground
...
4
...
For a point C on the
opposite bank, ∠ BAC = 56◦ and ∠ ABC = 41◦ , as in the picture
on the right
...
)
5
...
The top of the tower has angles of elevation α and β from A
and B, respectively, as in the picture on the right
...
Assuming that both sides of
the river are at the same elevation, show that the height h of
the tower is
h =
d
(cot α)2
+ (cot β)2
500
A
56◦
B
41◦
w
C
N
E
h
β
α
...
The equatorial parallax of the moon has been observed to be approximately 57′
...
6 miles, estimate the distance from the center of the earth to the moon
...
15
...
An observer on earth measures an angle of 31′ 7′′ from one visible edge of the moon to the other
(opposite) edge
...
(Hint: Use Exercise 6 and see Example
1
...
)
Applications and Solving Right Triangles • Section 1
...
A ball bearing sits between two metal grooves, with the top groove having an
angle of 120◦ and the bottom groove having an angle of 90◦ , as in the picture
on the right
...
21
120◦
1 ′′
2
90◦
1
...
The machine tool diagram on the right shows a symmetric worm
thread, in which a circular roller of diameter 1
...
Find the amount d that the top of the roller rises above the
top of the thread, given the information in the diagram
...
)
d
1
...
Repeat Exercise 9 using 1
...
11
...
For 0◦ < θ < 90◦ in the slider-crank mechanism in Example 1
...
(Hint: In Figure 1
...
2 draw line segments from B perpendicular to O A and AC
...
In the figure on the right, ∠ BAC = θ and BC = a
...
(Hint: What is the angle ∠ ACD ?)
1
28
54◦
′′
′′
r
11
2
13
...
In this view, the rounded tip is part of a circle of radius r , and the slanted
sides are tangent to that circle and form an angle of 54◦
...
Use the information in the
diagram to find the radius r
...
3
...
15
...
c = 6, B = 35◦
19
...
b = 1, c = 2
21
...
a = 2, B = 8◦
c
a
17
...
a = 2, c = 7
B
A
b
C
23
...
3
...
3
24
...
10 in Section 1
...
For example, we showed that cot 75◦ = 6− 2
...
We will now describe another method for finding the
6+ 2
exact values of the trigonometric functions of 15◦
...
The method
2
is illustrated in Figure 1
...
5 and is described below
...
5◦
C
15◦
B
30◦
A
60◦
O
Q
1
Figure 1
...
5
Draw a semicircle of radius 1 centered at a point O on a horizontal line
...
3
...
Draw
a line straight down from P to the horizontal line at the point Q
...
Then create a third semicircle in the same way: Let B be the left
endpoint of the second semicircle, then draw a new semicircle centered at B with radius equal to
BP
...
In general, it can be
shown that the line segment from the center of the new semicircle to P makes an angle with the
horizontal line equal to half the angle from the previous semicircle’s center to P
...
(Hint: What is the supplement of 60◦ ?)
(b) Explain why ∠ PBQ = 15◦ and ∠ PCQ = 7
...
(c) Use Figure 1
...
5 to find the exact values of sin 15◦ , cos 15◦ , and tan 15◦
...
)
(d) Use Figure 1
...
5 to calculate the exact value of tan 7
...
(e) Use the same method but with an initial angle of ∠ POQ = 45◦ to find the exact values of
sin 22
...
5◦ , and tan 22
...
Applications and Solving Right Triangles • Section 1
...
A manufacturer needs to place ten identical ball bearings against
the inner side of a circular container such that each ball bearing touches two other ball bearings, as in the picture on the
right
...
23
r
4
(a) Find the common radius r of the ball bearings
...
What is the largest possible
(outer) radius of the ring such that it is not on top
of the ball bearings and its base is level with the
base of the container?
26
...
That is, each of the eight
sides is tangent to the circle, as in the picture on the right
...
(b) If you were to increase the number of sides of the
polygon, would the area inside it increase or decrease?
What number would the area approach, if any? Explain
...
That is,
draw a regular octagon such that each of its eight vertexes
touches the circle
...
27
...
A
(a) Find the length of the diagonal line segment AB
...
B
B
28
...
3
...
Show that:
AD
(a) BC =
cot α − cot β
(b) AC =
AD · tan β
tan β − tan α
AD · sin α
(c) BD =
sin (β − α)
(Hint: What is the measure of the angle ∠ ABD ?)
α
A
β
D
C
Figure 1
...
6
29
...
How
many miles farther out is Person B’s horizon than Person A’s? (Note: 1 mile = 5280 ft)
24
Chapter 1 • Right Triangle Trigonometry
§1
...
4 Trigonometric Functions of Any Angle
To define the trigonometric functions of any angle - including angles less than 0◦ or greater
than 360◦ - we need a more general definition of an angle
...
The ray O A is called the initial side of the angle, and OB
is the terminal side of the angle (see Figure 1
...
1(a))
...
4
...
If
the rotation is counter-clockwise then we say that the angle is positive, and the angle is
negative if the rotation is clockwise (see Figure 1
...
1(b))
...
6 Not rotating O A constitutes an angle of 0◦
...
For example, notice that 30◦ and 390◦ have the same
terminal side in Figure 1
...
2, since 30 + 360 = 390
...
4
...
It is often assumed that the number 360 was used because the Babylonians (supposedly) thought that there
were 360 days in a year (a year, of course, is one full revolution of the Earth around the Sun)
...
e
...
The Babylonian mile was large enough
(approximately 7 of our miles) to be divided into 30 equal parts for convenience, thus giving 12 × 30 = 360 equal
parts in a full rotation
...
26 in H
...
, New York:
Saunders College Publishing, 1983
...
4
25
We can now define the trigonometric functions of any angle in terms of Cartesian coordinates
...
The first number, x, is the point’s x coordinate, and the second number, y, is
its y coordinate
...
This divides the
x y-coordinate plane into four quadrants (denoted by QI, QII, QIII, QIV), based on the signs
of x and y (see Figure 1
...
3(a)-(b))
...
4
...
We say that θ is in standard position if its initial side is the
positive x-axis and its vertex is the origin (0, 0)
...
4
...
(Note that r = x2 + y2
...
2)
csc θ =
r
y
sec θ =
r
x
cot θ =
x
y
(1
...
e
...
Also, notice
that | sin θ | ≤ 1 and | cos θ | ≤ 1, since | y | ≤ r and | x | ≤ r in the above definitions
...
4
Notice that in the case of an acute angle these definitions are equivalent to our earlier
definitions in terms of right triangles: draw a right triangle with angle θ such that x =
adjacent side, y = opposite side, and r = hypotenuse
...
4
...
y
y
hy
p
r
ot
en
us
e
( x, y)
y
opposite side
θ
180◦
x
0
90◦
QII
◦
90 < θ < 180◦
0◦
0
x
QIV
270 < θ < 360◦
QIII
180 < θ < 270◦
◦
◦
x
adjacent side
QI
0 < θ < 90◦
◦
270◦
(a) Acute angle θ
(b) Angles by quadrant
Figure 1
...
4
In Figure 1
...
4(b) we see in which quadrants or on which axes the terminal side of an angle
0 ≤ θ < 360◦ may fall
...
4
...
2) and (1
...
For example, sin θ < 0 when y < 0
...
4
...
4
...
4
27
Example 1
...
◦
◦
Solution: We know 120 = 180 − 60
...
7 in Section 1
...
Drawing that triangle in QII so that the hypotenuse is on the terminal
side of 120◦ makes r = 2, x = −1, and y = 3
...
21
Find the exact values of all six trigonometric functions of 225◦
...
By Example 1
...
2, we see that we can use the point (−1, −1) on the terminal side of
the angle 225◦ in QIII, since we saw in that example that a basic right
triangle with a 45◦ angle has adjacent side of length 1, opposite side
of length 1, and hypotenuse of length 2, as in the figure on the right
...
Hence:
y
−1
sin 225 =
=
r
2
x
−1
cos 225 =
=
r
2
◦
◦
csc 225◦ =
r
=− 2
y
sec 225◦ =
225◦
1
x
0
◦
45
1
2
(−1, −1)
y
−1
tan 225 =
=
=1
x
−1
r
=− 2
x
◦
cot 225◦ =
x
−1
=
=1
y
−1
Example 1
...
◦
◦
y
◦
Solution: We know that 330 = 360 − 30
...
7 in Section
1
...
Drawing that triangle in QIV so that the hypotenuse is on the
terminal side of 330◦ makes r = 2, x = 3, and y = −1
...
4
Example 1
...
y
(0, 1) 90◦
Solution: These angles are different from the angles we have considered so far, in that the terminal sides lie along either the x-axis
or the y-axis
...
However, the values of the trigonometric
x
0
(1, 0)
functions are easy to calculate by picking the simplest points on their (−1, 0)
terminal sides and then using the definitions in formulas (1
...
3)
...
4
...
You could think of the
Figure 1
...
6
line segment from the origin to the point (1, 0) as sort of a degenerate
right triangle whose height is 0 and whose hypotenuse and base have
the same length 1
...
Hence:
0
y
=
=0
r
1
cos 0◦ =
x
1
=
=1
r
1
tan 0◦ =
y
0
=
=0
x
1
r
1
=
= undefined
y
0
sec 0◦ =
r
1
=
=1
x
1
cot 0◦ =
x
1
=
= undefined
y
0
sin 0◦ =
csc 0◦ =
Note that csc 0◦ and cot 0◦ are undefined, since division by 0 is not allowed
...
4
...
Again, you could think of the line segment from the origin to (0, 1) as a degenerate right
triangle whose base has length 0 and whose height equals the length of the hypotenuse
...
Hence:
sin 180◦ =
csc 180◦ =
0
y
=
=0
r
1
1
r
=
= undefined
y
0
cos 180◦ =
x
−1
=
= −1
r
1
sec 180◦ =
r
1
=
= −1
x
−1
tan 180◦ =
cot 180◦ =
y
0
=
=0
x
−1
x
−1
=
= undefined
y
0
Lastly, for 270◦ use the point (0, −1) so that r = 1, x = 0, and y = −1
...
4
29
The following table summarizes the values of the trigonometric functions of angles between 0◦ and 360◦ which are integer multiples of 30◦ or 45◦ :
Table 1
...
For example, sin 360◦ = sin 0◦ , cos 390◦ = cos 30◦ , tan 540◦ = tan 180◦ , sin (−45◦ ) =
sin 315◦ , etc
...
Angles such as these, which have the
same initial and terminal sides, are called coterminal
...
20-1
...
That
acute angle has a special name: if θ is a nonacute angle then we say that the reference
angle for θ is the acute angle formed by the terminal side of θ and either the positive or
30
Chapter 1 • Right Triangle Trigonometry
§1
...
So in Example 1
...
21, 45◦ is the reference angle for θ = 225◦ ; and in Example
1
...
Example 1
...
y
(a) Which angle between 0◦ and 360◦ has the same terminal side
(and hence the same trigonometric function values) as θ ?
208◦
(b) What is the reference angle for θ ?
◦
◦
0
◦
Solution: (a) Since 928 = 2 × 360 + 208 , then θ has the same terminal side as 208◦ , as in Figure 1
...
7
...
28◦
x
928◦
Figure 1
...
7
Example 1
...
Find the exact values of sin θ and tan θ
...
8 in Section 1
...
That is,
adjacent
4
draw a right triangle and interpret cos θ as the ratio hypotenuse of two of its sides
...
By the Pythagorean
Theorem, the length of the opposite side must then be 3
...
4
...
Thus, we have two possibilities, as shown in Figure 1
...
8
below:
y
y
(−4, 3)
θ
4
5
θ
3
4
0
x
0
x
3
5
(−4, −3)
(a) θ in QII
Figure 1
...
8
(b) θ in QIII
4
cos θ = − 5
When θ is in QII, we see from Figure 1
...
8(a) that the point (−4, 3) is on the terminal side of θ , and
y
y
3
3
so we have x = −4, y = 3, and r = 5
...
When θ is in QIII, we see from Figure 1
...
8(b) that the point (−4, −3) is on the terminal side of θ , and
y
y
3
so we have x = −4, y = −3, and r = 5
...
5
−4
Thus, either sin θ =
3
5
3
3
and tan θ = − 4 or sin θ = − 5 and tan θ =
3
4
...
So it suffices to remember the
signs of sin θ , cos θ , and tan θ :
Trigonometric Functions of Any Angle • Section 1
...
1
...
−127◦
3
...
−313◦
5
...
621◦
7
...
2009◦
9
...
−514◦
11
...
In which quadrant(s) do sine and cosine have the opposite sign?
13
...
In which quadrant(s) do sine and tangent have the opposite sign?
15
...
In which quadrant(s) do cosine and tangent have the opposite sign?
For Exercises 17-21, find the reference angle for the given angle
...
317◦
18
...
−126◦
20
...
275◦
For Exercises 22-26, find the exact values of sin θ and tan θ when cos θ has the indicated value
...
cos θ =
1
2
23
...
cos θ = 0
25
...
cos θ = 1
For Exercises 27-31, find the exact values of cos θ and tan θ when sin θ has the indicated value
...
sin θ =
1
2
1
28
...
sin θ = 0
2
30
...
sin θ = 1
For Exercises 32-36, find the exact values of sin θ and cos θ when tan θ has the indicated value
...
tan θ =
1
2
1
33
...
tan θ = 0
35
...
tan θ = 1
For Exercises 37-40, use Table 1
...
37
...
Does tan 300◦ − tan 30◦ = tan 270◦ ?
39
...
Does cos 240◦ = (cos 120◦ )2 − (sin 120◦ )2 ?
41
...
3 to include all integer multiples of 15◦
...
10 in Section 1
...
32
Chapter 1 • Right Triangle Trigonometry
§1
...
5 Rotations and Reflections of Angles
Now that we know how to deal with angles of any measure, we will take a look at how certain
geometric operations can help simplify the use of trigonometric functions of any angle, and
how some basic relations between those functions can be made
...
To rotate an angle means to rotate its terminal side around the origin when the angle
is in standard position
...
In Figure 1
...
1 we see an angle θ in QI which is rotated
by 90◦ , resulting in the angle θ + 90◦ in QII
...
This forces the other angle of the right triangle in QII
to be θ
...
5
...
The rotation of θ by 90◦ does not change the length r of its terminal
side, so the hypotenuses of the similar right triangles are equal, and hence by similarity
the remaining corresponding sides are also equal
...
5
...
Hence, by definition,
sin (θ + 90◦ ) =
x
−y
x
= cos θ , cos (θ + 90◦ ) =
= − sin θ , tan (θ + 90◦ ) =
= − cot θ
...
In general, the following relations hold for all angles θ :
sin (θ + 90◦ ) = cos θ
(1
...
5)
tan (θ + 90◦ ) = − cot θ
(1
...
5
33
Example 1
...
e
...
5
...
We will show that the slopes of perpendicular lines are negative reciprocals
...
5
...
y
y = mx + b
y
y = m2 x + b2
rise
b
m=
run
0
rise
run
x
0
y = m1 x + b1
b2
x
b1
(a) Slope of a line
(b) Perpendicular lines
Figure 1
...
2
First, suppose that a line y = mx + b has nonzero slope
...
5
...
For m > 0 we see that θ is acute and tan θ = rise = m
...
5
...
Since the run is always positive,
our definition of tan θ from Section 1
...
5
...
Hence:
For a line y = mx + b with m = 0, the slope is given by m = tan θ , where θ is the angle formed by
the positive x-axis and the part of the line above the x-axis
...
5
...
So if θ is the angle that the line y = m 1 x + b 1 makes with the positive x-axis, then θ + 90◦ is
the angle that the line y = m 2 x + b 2 makes with the positive x-axis
...
But by formula (1
...
Hence,
1
1
m 2 = − cot θ = − tan θ = − m1
...
5
Rotating an angle θ by 90◦ in the clockwise direction results in the angle θ − 90◦
...
4)−(1
...
Since (θ − 90◦ ) + 90◦ = θ , this gives us:
sin (θ − 90◦ ) = − cos θ
(1
...
8)
tan (θ − 90◦ ) = − cot θ
(1
...
Notice from Figure 1
...
4 that the angles
θ ± 180◦ have the same terminal side, and are in the quadrant opposite θ
...
5
...
10)
(1
...
12)
A reflection is simply the mirror image of an object
...
5
...
Notice that if we
first reflect the object in QI around the y-axis and then follow that with
a reflection around the x-axis, we get an image in QIII
...
Notice also that a reflection around the
y-axis is equivalent to a reflection around the x-axis followed by a rotation
of 180◦ around the origin
...
5
...
5
35
Applying this to angles, we see that the reflection of an angle θ around the x-axis is the
angle −θ , as in Figure 1
...
6
...
5
...
Hence:
sin (−θ ) = − sin θ
(1
...
14)
tan (−θ ) = − tan θ
(1
...
14) because it depends on x,
and not on y, for a point ( x, y) on the terminal side of θ
...
Thus, the cosine function is even, while the sine and
tangent functions are odd
...
4)−(1
...
13)−(1
...
16)
cos (90◦ − θ ) = sin θ
(1
...
18)
Note that formulas (1
...
18) extend the Cofunction Theorem from Section 1
...
Similarly, formulas (1
...
12) and (1
...
15) give:
sin (180◦ − θ ) = sin θ
(1
...
20)
tan (180◦ − θ ) = − tan θ
(1
...
5
Notice that reflection around the y-axis is equivalent to reflection around the x-axis (θ →
−θ ) followed by a rotation of 180◦ (−θ → −θ + 180◦ = 180◦ − θ ), as in Figure 1
...
7
...
5
...
However, there are
two reasons for why they are useful
...
Second, they can help in determining which angles have a given trigonometric
function value
...
27
Find all angles 0◦ ≤ θ < 360◦ such that sin θ = −0
...
✄
sin
Solution: Using the ✂ −1 ✁button on a calculator with −0
...
7 Since θ = −43◦ is in QIV, its reflection 180◦ − θ around the y-axis will be
in QIII and have the same sine value
...
5
...
Also, we
know that −43◦ and −43◦ + 360◦ = 317◦ have the same trigonometric function values
...
682 are θ = 223◦ and 317◦
...
5
...
✂
✁
Rotations and Reflections of Angles • Section 1
...
Let θ = 32◦
...
Repeat Exercise 1 with θ = 248◦
...
Repeat Exercise 1 with θ = −248◦
...
We proved formulas (1
...
6) for any angle θ in QI
...
5
...
4)-(1
...
e
...
6
...
26 we used the formulas involving θ + 90◦ to prove that the slopes of perpendicular lines are negative reciprocals
...
(Hint: Only the last paragraph in that example needs to be modified
...
sin θ = 0
...
sin θ = 0
...
cos θ = 0
...
sin θ = 0
11
...
7813
12
...
6294
13
...
9816
14
...
514
15
...
2, we
claimed that in a right triangle △ ABC it was possible to draw
a line segment CD from the right angle vertex C to a point D
on the hypotenuse AB such that CD ⊥ AB
...
(Hint: Notice how △ ABC is placed
on the x y-coordinate plane
...
y
B (0, b)
C (0, 0)
D
A (a, 0)
x
16
...
Let y = m 1 x + b 1 and y = m 2 x + b 2 be perpendicular lines (with nonzero
1
slopes), as in the picture below
...
(Hint: Think of similar triangles and the definition of slope
...
Prove formulas (1
...
21) by using formulas (1
...
12) and (1
...
15)
...
3 we saw how to solve a right triangle: given two sides, or one side and one
acute angle, we could find the remaining sides and angles
...
For a general triangle, which may or may not have a right angle, we will again need three
pieces of information
...
In this chapter we will learn how to solve a general triangle in all four of the above cases
...
There are two types
of oblique triangles: an acute triangle has all acute angles, and an obtuse triangle has
one obtuse angle
...
2
...
1
...
sin A
sin B
sin C
(2
...
1) can be written as
sin A
sin B
sin C
=
=
,
a
b
c
(2
...
3)
The Law of Sines • Section 2
...
To prove the Law of Sines, let △ ABC be an oblique triangle
...
1
...
1
...
In each case, draw the altitude1
from the vertex at C to the side AB
...
1
...
1
...
C
b
A
h
C
a
c
b
a
B
A
(a) Acute triangle
c
B
h
180◦ − B
(b) Obtuse triangle
Figure 2
...
1 Proof of the Law of Sines for an oblique triangle △ ABC
Let h be the height of the altitude
...
1
...
4)
h
= sin B
a
(2
...
1
...
19) in Section 1
...
Thus, solving
a
for h in equation (2
...
4) gives
a sin B
= sin A ,
b
(2
...
sin A
sin B
(2
...
8)
sin B
sin C
so putting the last two equations together proves the theorem
...
1 Recall from geometry that an altitude of a triangle is a perpendicular line segment from any vertex to the line
containing the side opposite the vertex
...
1
Example 2
...
Solve the triangle △ ABC given a = 10, A = 41◦ , and C = 75◦
...
In this
example we need to find B, b, and c
...
So by the Law of Sines we have
b
a
=
sin B
sin A
⇒
b =
a sin B
10 sin 64◦
=
sin A
sin 41◦
⇒
b = 13
...
7
...
2
Case 2: Two sides and one opposite angle
...
C
b = 30
Solution: In this example we know the side a and its opposite angle A ,
and we know the side b
...
By the Law of Sines, we have
sin B
sin A
=
b
a
✄
⇒
sin B =
b sin A
30 sin 25◦
=
a
18
⇒
a = 18
c
B
sin B = 0
...
sin
Using the ✂ −1 ✁ button on a calculator gives B = 44
...
However, recall from Section 1
...
So there is a second possible solution for B, namely 180◦ − 44
...
2◦
...
8◦ and once for B = 135
...
8◦
C = 180◦ − A − B = 180◦ − 25◦ − 44
...
2◦
c
a
a sin C 18 sin 110
...
2◦
C = 180◦ − A − B = 180◦ − 25◦ − 135
...
8◦
c
a
a sin C 18 sin 19
...
4
Hence, B = 44
...
2◦ , c = 40 and B = 135
...
8◦ , c = 14
...
This means that there are two possible triangles, as shown in Figure 2
...
2
...
2◦
b = 30
a = 18
◦
A = 25
c = 40
b = 30
◦
B = 44
...
8◦
Figure 2
...
2
◦
A = 25
C = 19
...
4
a = 18
B = 135
...
2◦
Two possible solutions
The Law of Sines • Section 2
...
2 we saw what is known as the ambiguous case
...
It is also possible for there to be exactly one solution or no solution at all
...
3
Case 2: Two sides and one opposite angle
...
Solution: By the Law of Sines, we have
sin A
sin B
=
b
a
⇒
sin B =
b sin A
12 sin 30◦
=
a
5
⇒
sin B = 1
...
Thus, there is no solution
...
For a triangle
△ ABC , suppose that we know the sides a and b and the angle A
...
1
...
C
b
A
C
C
a
b
h
B
c
A
(a) a < h: No solution
b
h a
a
a
B
A
(b) a = h: One solution
B
(c) h < a < b: Two solutions
C
b
B
h
a
b
A
c
B
(d) a ≥ b: One solution
Figure 2
...
3
The ambiguous case when A is acute
If A is acute, then the altitude from C to AB has height h = b sin A
...
1
...
3); there
is exactly one solution - namely, a right triangle - when a = h; and there are two solutions
when h < a < b (as was the case in Example 2
...
When a ≥ b there is only one solution, even
though it appears from Figure 2
...
3(d) that there may be two solutions, since the dashed arc
intersects the horizontal line at two points
...
1
...
If A is not acute (i
...
A is obtuse or a right angle), then the situation is simpler: there is
no solution if a ≤ b, and there is exactly one solution if a > b (see Figure 2
...
4)
...
1
C
a
a
b
b
B
A
(a) a ≤ b: No solution
Figure 2
...
4
A
B
(b) a > b: One solution
The ambiguous case when A ≥ 90◦
Table 2
...
Of
course, the letters can be interchanged, e
...
replace a and A by c and C , etc
...
1
Summary of the ambiguous case
0◦ < A < 90◦
a < b sin A : No solution
a = b sin A : One solution
b sin A < a < b : Two solutions
a ≥ b : One solution
90◦ ≤ A < 180◦
a ≤ b : No solution
a > b : One solution
There is an interesting geometric consequence of the Law of Sines
...
1
that in a right triangle the hypotenuse is the largest side
...
What
the Law of Sines does is generalize this to any triangle:
In any triangle, the largest side is opposite the largest angle
...
If C = 90◦ then we already
know that its opposite side c is the largest side
...
In both cases, we have A ≤ C and B ≤ C
...
If C is acute, then A and B are also acute
...
1
...
If we pick points
( x1 , y1 ) and ( x2 , y2 ) on the terminal sides of A and C , respectively,
so that their distance to the origin is the same number r , then we
see from the picture that y1 ≤ y2 , and hence
sin A =
y1
y2
≤
= sin C
...
1
...
Thus, sin A ≤ sin C and sin B ≤
sin C when C is acute
...
The Law of Sines • Section 2
...
If A > 180◦ − C then A + C > 180◦ ,
which is impossible
...
Likewise, B ≤ 180◦ − C
...
But we know from Section 1
...
Hence, sin A ≤ sin C and sin B ≤
sin C when C is obtuse
...
c
sin C
sin C
c
By a similar argument, b ≤ c
...
e
...
QED
Exercises
For Exercises 1-9, solve the triangle △ ABC
...
a = 10, A = 35◦ , B = 25◦
2
...
A = 40◦ , B = 45◦ , c = 15
4
...
a = 40, A = 25◦ , c = 30
6
...
a = 12, A = 94◦ , b = 15
8
...
a = 22, A = 50◦ , c = 27
10
...
Use a ruler and
a protractor to measure the sides a, b, c and the angles A , B, C of the triangle
...
Verify this for your triangle
...
An observer on the ground measures an angle of inclination of 30◦ to an approaching airplane,
and 10 seconds later measures an angle of inclination of 55◦
...
(Note: 1 mile = 5280 ft)
10 seconds pass
✈
6000 ft
✈
55◦
30◦
12
...
(Hint: One of the angles is known
...
c
sin C
a
sin (B + C )
14
...
c
sin C
15
...
Find the lengths of the sides
...
For a triangle △ ABC , show that
16
...
44
Chapter 2 • General Triangles
§2
...
2 The Law of Cosines
We will now discuss how to solve a triangle in Case 3: two sides and the angle between them
...
Example 2
...
Solve the triangle △ ABC given A = 30◦ , b = 4, and c = 5
...
For
4
5
example, to solve for a we could use the equation sin B = sin C to solve for sin B in terms of sin C and
4
a
substitute that into the equation sin 30◦ = sin B
...
To solve the triangle in the above example, we can use the Law of Cosines:
Theorem 2
...
Law of Cosines: If a triangle has sides of lengths a, b, and c opposite the
angles A , B, and C , respectively, then
a2 = b2 + c2 − 2 bc cos A ,
(2
...
10)
2
2
2
c = a + b − 2ab cos C
...
11)
To prove the Law of Cosines, let △ ABC be an oblique triangle
...
2
...
2
...
In each case, draw the altitude
from the vertex at C to the side AB
...
2
...
2
...
Let h be the height of the altitude
...
2
...
2
45
For each triangle in Figure 2
...
1, we see by the Pythagorean Theorem that
h2 = a2 − x2
(2
...
2
...
(2
...
12) into equation (2
...
But we see from Figure 2
...
1(a) that x = a cos B, so
b2 = a2 + c2 − 2 ca cos B
...
14)
And for the obtuse triangle in Figure 2
...
1(b) we see that
b2 = h2 + ( c + x)2
...
15)
Thus, substituting the expression for h2 in equation (2
...
15) gives
b2 = a2 − x2 + ( c + x)2
= a2 − x2 + c2 + 2 cx + x2
= a2 + c2 + 2 cx
...
2
...
5 that
cos (180◦ − B) = − cos B
...
(2
...
10) in the Law of Cosines
...
By similar arguments for A and C we get
the other two formulas
...
The
Law of Cosines can be viewed as a generalization of the Pythagorean Theorem
...
9)-(2
...
That is, replace a
by b, replace b by c, and replace c by a (likewise for the capital letters)
...
46
Chapter 2 • General Triangles
§2
...
Notice in
the Law of Cosines that if two sides and their included angle are known (e
...
b, c, and A ),
then we have a formula for the square of the third side
...
4
...
5
Case 3: Two sides and the angle between them
...
C
Solution: We will use the Law of Cosines to find a, use it again to find
B, then use C = 180◦ − A − B
...
36
⇒
b=4
A = 30◦
a
c=5
B
a = 2
...
2
Now we use the formula for b to find B:
b2 = c2 + a2 − 2 ca cos B
⇒
⇒
⇒
c 2 + a2 − b 2
2 ca
52 + (2
...
6091
cos B =
2(5)(2
...
5◦
Thus, C = 180◦ − A − B = 180◦ − 30◦ − 52
...
5◦
...
5 that there was only one solution
...
The reason is simple: when joining two line segments at a common vertex to form
an angle, there is exactly one way to connect their free endpoints with a third line segment,
regardless of the size of the angle
...
5, to
find the angle B
...
This is in contrast
to using the sine function; as we saw in Section 2
...
To see this, suppose that we had used the Law of Sines to find B in Example 2
...
7937 ⇒ B = 52
...
5◦
a
2
...
5◦
as too large, since it would make A + B = 157
...
5◦ , which seems like
it could be a valid solution
...
Why? Because the largest
side in the triangle is c = 5, which (as we learned in Section 2
...
But C = 22
...
sin B =
The Law of Cosines • Section 2
...
e
...
We will now see how to use
the Law of Cosines for that case
...
6
Case 4: Three sides
...
C
Solution: We will use the Law of Cosines to find B and C , then use A =
180◦ − B − C
...
6875
2(4)(2)
a=2
c=4
B
cos B =
B = 46
...
25
2(2)(3)
cos C =
C = 104
...
6◦ − 104
...
9◦
...
Example 2
...
Solve the triangle △ ABC given a = 2, b = 3, and c = 6
...
139 ,
2 bc
2(3)(6)
a=2
c=6
B
which is impossible since | cos A | ≤ 1
...
We could have saved ourselves some effort by recognizing that the length
of one of the sides ( c = 6) is greater than the sums of the lengths of the
remaining sides (a = 2 and b = 3), which (as the picture on the right shows)
is impossible in a triangle
...
The
following example gives an idea of how to do this
...
2
Example 2
...
Solve the triangle △ ABC given a = 18, A = 25◦ , and b = 30
...
2 from Section 2
...
8◦ , C = 110
...
2◦ , C = 19
...
4
...
38 c + 576 = 0 ,
which is a quadratic equation in c, so we know that it can have either zero, one, or two real roots
(corresponding to the number of solutions in Case 2)
...
38 ±
(54
...
4
...
For c = 40 we get
cos B =
c 2 + a2 − b 2
402 + 182 − 302
=
= 0
...
7◦
⇒
C = 110
...
The
other solution set can be obtained similarly
...
Example 2
...
Solution: Let a and b be the lengths of the sides, and let the diagonals opposite the angles C and D have lengths c and d , respectively,
as in Figure 2
...
2
...
By properties of parallelograms, we know that D = 180◦ − C , so
d 2 = a2 + b2 − 2ab cos (180◦ − C )
= a2 + b2 + 2ab cos C ,
since cos (180◦ − C ) = − cos C
...
= 2 ( a2 + b 2 )
...
2
...
2
49
Exercises
For Exercises 1-6, solve the triangle △ ABC
...
A = 60◦ , b = 8, c = 12
2
...
a = 7, B = 60◦ , c = 9
4
...
a = 6, b = 4, c = 1
6
...
The diagonals of a parallelogram intersect at a 42◦ angle and have lengths of 12 and 7 cm
...
(Hint: The diagonals bisect each other
...
Two trains leave the same train station at the same time, moving along straight tracks that form
a 35◦ angle
...
Three circles with radii of 4, 5, and 6 cm, respectively, are tangent to each other externally
...
10
...
2
...
60◦
3
...
5
Figure 2
...
3
Exercise 10
Figure 2
...
4 Exercise 11
11
...
At either point of intersection, the tangent lines to the circles form a 60◦ angle, as in Figure 2
...
4 above
...
12
...
13
...
a
b
c
2abc
14
...
2abc
What do the terms in parentheses represent geometrically? Use your answer to explain why
cos A + cos B + cos C > 0 for any triangle, even if one of the cosines is negative
...
Prove the Law of Cosines (i
...
formulas (2
...
11)) for right triangles
...
Recall from elementary geometry that a median of a triangle is a line segment from any vertex
to the midpoint of the opposite side
...
2 It turns out that 1 < cos A + cos B + cos C ≤ 3/2 for any triangle, as we will see later
...
2
17
...
Show that this
formula is equivalent to formula (2
...
18
...
Let r e be the radius of the earth, let r s be the distance from the center of the earth to the
satellite (called the orbital radius of the satellite), and let d be the distance from the earth station
to the satellite
...
2
...
satellite
✸
local horizontal
d
rs
E
earth station
❚
ψ
γ
re
center
of earth
Figure 2
...
5
Use the Law of Cosines to show that
d = rs
1+
re
rs
2
−2
re
cos γ ,
rs
and then use E = ψ − 90◦ and the Law of Sines to show that
cos E =
sin γ
re
1+
rs
2
...
3
3 See pp
...
P RATT AND C
...
B OSTIAN, Satellite Communications, New York: John Wiley & Sons,
1986
...
3
51
2
...
An alternative to the Law of Cosines for Case 3 (two sides and the included angle)
is the Law of Tangents:
Theorem 2
...
Law of Tangents: If a triangle has sides of lengths a, b, and c opposite the
angles A , B, and C , respectively, then
tan 1 ( A − B)
a−b
2
=
,
1
a+b
tan 2 ( A + B)
(2
...
18)
tan 1 (C − A )
c−a
2
...
19)
Note that since tan (−θ ) = − tan θ for any angle θ , we can switch the order of the letters in
each of the above formulas
...
17) as
tan 1 (B − A )
b−a
2
,
=
1
b+a
tan 2 (B + A )
(2
...
If a > b, then it is usually more convenient to use
formula (2
...
20) is more convenient when b > a
...
10
Case 3: Two sides and the included angle
...
Solution: A + B + C = 180◦ , so A + B = 180◦ − C = 180◦ − 96◦ = 84◦
...
2251
◦
1
⇒ A − B = 25
...
7
...
4◦
A + B = 84◦
−− − − − − − −
2A
= 109
...
7◦
⇒
B = 84◦ − 54
...
7◦
⇒
c = 6
...
3◦
a=5
B
52
Chapter 2 • General Triangles
§2
...
17) would be 0 (since tan 0◦ = 0)
...
For this reason, and perhaps also because of the
somewhat unusual way in which it is used, the Law of Tangents seems to have fallen out of
favor in trigonometry books lately
...
4
Related to the Law of Tangents are Mollweide’s equations:5
Mollweide’s equations: For any triangle △ ABC ,
sin 1 ( A − B)
a−b
2
, and
=
1
c
cos 2 C
(2
...
1
c
sin 2 C
(2
...
For this
reason, either equation can be used to check a solution of a triangle
...
Example 2
...
10
...
09, A = 54
...
3◦ , and C = 96◦
...
21):
sin 1 ( A − B)
a−b
2
=
c
cos 1 C
2
sin 1 (54
...
3◦ )
5−3
2
=
6
...
7◦
2
=
6
...
3284 = 0
...
0001) is due to rounding errors from the original solution, so we can conclude that both sides of the equation agree, and hence the solution is correct
...
In those days,
computations with large numbers were handled by taking logarithms and looking up values in a logarithm table
...
5 Named after the German astronomer and mathematician Karl Mollweide (1774-1825)
...
3
53
Example 2
...
In this
case all those conditions hold
...
22):
cos 1 ( A − B)
a+b
2
=
1
c
sin 2 C
1
cos 2 (55◦ − 60◦ )
6+7
=
9
sin 1 (65◦ )
2
cos (−2
...
5◦
1
...
86 ✗
Here the difference is far too large, so we conclude that there is no triangle with these parts
...
6
Exercises
For Exercises 1-3, use the Law of Tangents to solve the triangle △ ABC
...
a = 12, b = 8, C = 60◦
2
...
a = 7, B = 60◦ , c = 9
For Exercises 4-6, check if it is possible for a triangle to have the given parts
...
a = 5, b = 7, c = 10, A = 27
...
5◦ , C = 111
...
a = 3, b = 7, c = 9, A = 19
...
2◦ , C = 92
...
a = 6, b = 9, c = 9, A = 39◦ , B = 70
...
5◦
7
...
Show that tan 1 ( A − B) =
2
8
...
a− b
a+ b
...
(Hint: Draw the altitude from the vertex
c − a cos B
C to AB
...
9
...
For any triangle △ ABC , show that c = b cos A + a cos B
...
11
...
12
...
The quadrilateral
has eight parts: four sides and four angles
...
6 There are (complex) geometric proofs of the Law of Tangents and Mollweide’s equations
...
96-98 in P
...
R IDER, Plane and Spherical Trigonometry, New York: The Macmillan Company, 1942
...
4
2
...
We will now use that, combined with some trigonometry, to derive more formulas for
the area when given various parts of the triangle
...
Suppose that we have a triangle △ ABC , in which A can be either acute, a right angle, or
obtuse, as in Figure 2
...
1
...
C
b
h
c
A
C
a
C
a
b h
B
h
c
A
B
◦
(a) A acute
(b) A = 90
Figure 2
...
1
a
b
c
A
B
(c) A obtuse
Area of △ ABC
In each case we draw an altitude of height h from the vertex at C to AB, so that the area
1
(which we will denote by the letter K ) is given by K = 2 hc
...
4
...
4
...
We thus get the following formula:
Area = K =
1
2
bc sin A
(2
...
Similar
arguments for the angles B and C give us:
Area = K =
Area = K =
1
2
1
2
ac sin B
(2
...
25)
Notice that the height h does not appear explicitly in these formulas, although it is implicitly
there
...
Example 2
...
Solution: Using formula (2
...
53
C
b=5
A = 33◦
a
c=7
B
The Area of a Triangle • Section 2
...
Suppose that we have a triangle △ ABC in which one side, say, a, and all three angles are
known
...
24) we get:
Area = K =
a2 sin B sin C
2 sin A
(2
...
27)
Area = K =
(2
...
14
Find the area of the triangle △ ABC given
A = 115◦ , B = 25◦ , C = 40◦ , and a = 12
...
26), the area K is given by:
b
a = 12
A = 115◦ c
a2 sin B sin C
2 sin A
122 sin 25◦ sin 40◦
=
2 sin 115◦
K = 21
...
Suppose that we have a triangle △ ABC in which all three sides are known
...
e
...
Then the area K of the triangle is
Area = K =
s ( s − a) ( s − b ) ( s − c )
...
29)
To prove this, first remember that the area K is one-half the base times the height
...
4
...
Squaring both sides gives us
K 2 = 1 h2 c 2
...
30)
4
7 Note that this is equivalent to knowing just two angles and a side (why?)
...
10-70 A
...
56
Chapter 2 • General Triangles
§2
...
4
...
C
b
h
C
a
c D
A
h
B
D
(a) A acute
Figure 2
...
2
a
b
A
c
B
(b) A obtuse
Proof of Heron’s formula
By the Pythagorean Theorem, we see that h2 = b2 − ( AD )2
...
4
...
And in Figure 2
...
2(b) we see that AD = b cos (180◦ − A ) = − b cos A
...
(2
...
Thus,
substituting equation (2
...
30), we have
1
K 2 = 4 b2 c2 (1 + cos A ) (1 − cos A )
...
32)
By the Law of Cosines we know that
b 2 + c 2 − a2
2 bc + b2 + c2 − a2
( b + c)2 − a2
(( b + c) + a) (( b + c) − a)
=
=
=
2 bc
2 bc
2 bc
2 bc
( a + b + c ) ( b + c − a)
,
=
2 bc
1 + cos A = 1 +
and similarly
b 2 + c 2 − a2
2 bc − b2 − c2 + a2
a2 − ( b − c)2
(a − ( b − c)) (a + ( b − c))
=
=
=
2 bc
2 bc
2 bc
2 bc
(a − b + c) (a + b − c)
...
32), we have
( a + b + c ) ( b + c − a) ( a − b + c ) ( a + b − c )
·
2 bc
2 bc
a+b+ c b+ c−a a−b+ c a+b− c
=
·
·
·
,
2
2
2
2
K2 =
1
4
b2 c2
and since we defined s = 1 (a + b + c), we see that
2
K 2 = s ( s − a) ( s − b ) ( s − c ) ,
so upon taking square roots we get
K =
s ( s − a) ( s − b ) ( s − c )
...
4
57
Example 2
...
1
1
Solution: Using Heron’s formula with s = 2 (a + b + c) = 2 (5 + 4 + 7) = 8, the
area K is given by:
K =
=
C
b=4
A
s ( s − a) ( s − b ) ( s − c )
8 (8 − 5) (8 − 4) (8 − 7) =
96
⇒
a=5
c=7
B
K = 4 6 ≈ 9
...
Heron’s formula is useful for theoretical purposes (e
...
in deriving other formulas)
...
Example 2
...
9999979, and c = 0
...
1
Solution: To use Heron’s formula, we need to calculate s = 2 (a + b + c)
...
0000008, which has 14 digits
...
When we then
divide that rounded value for a + b + c by 2 to get s, some calculators (e
...
the TI-83 Plus) will give a
rounded down value of 1000000
...
Other calculators may give some other inaccurate answer, depending on how they store
values internally
...
99999999999895, i
...
it
is basically 1
...
9 Luckily
there is a better formula10 for the area of a triangle when the three sides are known:
For a triangle △ ABC with sides a ≥ b ≥ c, the area is:
Area = K =
1
4
(a + ( b + c)) ( c − (a − b)) ( c + (a − b)) (a + ( b − c))
(2
...
Then perform the
operations inside the square root in the exact order in which they appear in the formula,
including the use of parentheses
...
For the triangle
in Example 2
...
What is amazing about this formula is that it
is just Heron’s formula rewritten! The use of parentheses is what forces the correct order of
operations for numerical stability
...
There is an excellent overview of this important subject in the
article What Every Computer Scientist Should Know About Floating-Point Arithmetic by D
...
sun
...
html
10 Due to W
...
eecs
...
edu/~wkahan/Triangle
...
4
Another formula11 for the area of a triangle given its three sides is given below:
For a triangle △ ABC with sides a ≥ b ≥ c, the area is:
Area = K =
1
2
a2 + c 2 − b 2
2
a2 c 2 −
2
(2
...
16, the above formula gives an answer of exactly K = 1 on
the same TI-83 Plus calculator that failed with Heron’s formula
...
1
...
a = 10, B = 95◦ , c = 35
3
...
A = 171◦ , B = 1◦ , C = 8◦ , b = 2
5
...
a = 5, b = 6, c = 5
7
...
4
...
C
3
...
5
Figure 2
...
3
θ
B
6
D
A
Exercise 7
Figure 2
...
4 Exercise 8
8
...
4
...
Show that the area K of ABCD is equal to half the product of its diagonals and the sine of
1
the angle they form, i
...
K = 2 AC · BD sin θ
...
From formula (2
...
Show that the triangle area formula
Area = K =
1
4
(a + ( b + c)) ( c − (a − b)) ( c + (a − b)) (a + ( b − c))
1
is equivalent to Heron’s formula
...
)
11
...
34) is equivalent to Heron’s formula
...
)
12
...
16, then use formula (2
...
Did it work?
11 Due to the Chinese mathematician Qiu Jiushao (ca
...
Circumscribed and Inscribed Circles • Section 2
...
5 Circumscribed and Inscribed Circles
Recall from the Law of Sines that any triangle △ ABC has a common ratio of sides to sines
of opposite angles, namely
a
b
c
=
=
...
e
...
Before proving this, we need to review some elementary geometry
...
In Figure 2
...
1(a),
∠ O is a central angle and we say that it intercepts the arc BC
...
5
...
In
Figure 2
...
1(b), ∠ A is an inscribed angle that intercepts the arc BC
...
4
...
Thus, inscribed angles which intercept the same arc are equal
...
5
...
2
We will now prove our assertion about the common ratio in the Law of Sines:
Theorem 2
...
For any triangle △ ABC , the radius R of its circumscribed circle is given by:
2R =
a
b
c
=
=
sin A
sin B
sin C
(Note: For a circle of diameter 1, this means a = sin A , b = sin B, and c = sin C
...
210-211 in R
...
AVERY, Plane Geometry, Boston: Allyn & Bacon, 1950
...
35)
60
Chapter 2 • General Triangles
§2
...
Then
O can be either inside, outside, or on the triangle, as in Figure 2
...
2 below
...
C
C
C
a
b
b
R
A
c
2
A
a
O
c
2
R
c
2
a
b
B
R
O
R
D
D
c
2
c
A
R
O
B
R
B
(a) O inside △ ABC
(b) O outside △ ABC
Figure 2
...
2
(c) O on △ ABC
Circumscribed circle for △ ABC
The radii O A and OB have the same length R , so △ AOB is an isosceles triangle
...
c
1
So ∠ AOD = 2 ∠ AOB and AD = 2
...
4 that ∠ ACB = 2 ∠ AOB
...
So since C = ∠ ACB, we have
c
c
c
AD
= 2 =
⇒ 2R =
,
sin C = sin ∠ AOD =
OA
R
2R
sin C
so by the Law of Sines the result follows if O is inside or outside △ ABC
...
5
...
Then AB is
a diameter of the circle, so C = 90◦ by Thales’ Theorem
...
QED
Example 2
...
Solution: We know that △ ABC is a right triangle
...
5
...
Thus,
2R =
a
3
= 3 = 5
sin A
5
⇒
R = 2
...
Note that since R = 2
...
Thus, AB must be a diameter of the circle, and so the center O of the
circle is the midpoint of AB
...
5
...
6
...
e
...
Circumscribed and Inscribed Circles • Section 2
...
5 units from A along AB
...
Notice from the proof of Theorem 2
...
Similar arguments for the other sides would
show that O is on the perpendicular bisectors for those sides:
Corollary 2
...
For any triangle, the center of its circumscribed circle is the intersection of
the perpendicular bisectors of the sides
...
Pick the radius large enough so that the
arcs intersect at two points, as in Figure 2
...
4
...
For the circumscribed circle of a triangle, you need the perpendicular bisectors of only two of the sides; their intersection will be
the center of the circle
...
5
...
18
Find the radius R of the circumscribed circle for the triangle △ ABC from Example 2
...
2: a = 2, b = 3, and c = 4
...
Solution: In Example 2
...
9◦ , so 2 R =
a
sin A
=
2
sin 28
...
14, so R = 2
...
In Figure 2
...
5(a) we show how to draw △ ABC : use a ruler to draw the longest side AB of length
c = 4, then use a compass to draw arcs of radius 3 and 2 centered at A and B, respectively
...
C
C
3
A
2
c=4
B
A
O
B
(a) Drawing △ ABC
(b) Circumscribed circle
Figure 2
...
5
In Figure 2
...
5(b) we show how to draw the circumscribed circle: draw the perpendicular bisectors
of AB and AC ; their intersection is the center O of the circle
...
62
Chapter 2 • General Triangles
§2
...
5 can be used to derive another formula for the area of a triangle:
Theorem 2
...
For a triangle △ ABC , let K be its area and let R be the radius of its circumscribed circle
...
(2
...
5 we have
2R =
a
b
c
=
=
sin A
sin B
sin C
sin A =
⇒
a
b
c
, sin B =
, sin C =
...
26) from Section 2
...
8 with Heron’s formula for the area of a triangle, we get:
1
Corollary 2
...
For a triangle △ ABC , let s = 2 (a + b + c)
...
(2
...
e
...
5
...
C
E
F
a
b
O
r
A
D
Figure 2
...
6
c
B
Inscribed circle for △ ABC
Let r be the radius of the inscribed circle, and let D , E , and F be the points on AB, BC , and
AC , respectively, at which the circle is tangent
...
Thus, △ O AD and △ O AF are equivalent triangles, since they are right triangles with the
same hypotenuse O A and with corresponding legs OD and OF of the same length r
...
Similarly, OB bisects B and
OC bisects C
...
Circumscribed and Inscribed Circles • Section 2
...
5
...
Since O A bisects A ,
r
we see that tan 1 A = AD , and so r = AD · tan 1 A
...
Similarly, DB = EB and FC = CE
...
Hence, r = ( s − a) tan
1
2 A
...
10
...
Then the radius r of its inscribed
circle is
1
1
(2
...
2
We also see from Figure 2
...
6 that the area of the triangle △ AOB is
Area(△ AOB) =
1
2 base × height
1
2
=
cr
...
Thus, the area K of △ ABC is
2
K = Area(△ AOB) + Area(△ BOC ) + Area(△ AOC ) =
=
r =
1
2 (a + b + c) r
K
=
s
1
2
cr +
1
2
ar +
1
2
br
= sr , so by Heron’s formula we get
s ( s − a) ( s − b ) ( s − c )
=
s
s ( s − a) ( s − b ) ( s − c )
=
s2
( s − a) ( s − b ) ( s − c )
...
11
...
Then the radius r of its inscribed
circle is
K
( s − a) ( s − b ) ( s − c )
=
...
39)
r =
s
s
Recall from geometry how to bisect an angle: use a compass centered at the vertex to draw an arc that intersects the sides of the
angle at two points
...
5
...
The line through that point and
the vertex is the bisector of the angle
...
d
d
A
Figure 2
...
7
64
Chapter 2 • General Triangles
§2
...
19
Find the radius r of the inscribed circle for the triangle △ ABC from Example 2
...
2:
a = 2, b = 3, and c = 4
...
Solution: Using Theorem 2
...
12
Figure 2
...
8 shows how to draw the inscribed circle: draw the
bisectors of A and B, then at their intersection use a compass
to draw a circle of radius r = 5/12 ≈ 0
...
O
B
A
Figure 2
...
8
Exercises
For Exercises 1-6, find the radii R and r of the circumscribed and inscribed circles, respectively, of
the triangle △ ABC
...
a = 2, b = 4, c = 5
2
...
a = 5, b = 7, C = 40◦
4
...
a = 10, b = 11, c = 20
...
a = 5, b = 12, c = 13
For Exercises 7 and 8, draw the triangle △ ABC and its circumscribed and inscribed circles accurately, using a ruler and compass (or computer software)
...
a = 2 in, b = 4 in, c = 5 in
8
...
For any triangle △ ABC , let s = 2 (a + b + c)
...
s ( s − c)
10
...
11
...
2 (a + b + c)
12
...
(a) Find the exact value of the radius R of the circumscribed circle of △ ABC
...
(c) How much larger is R than r ?
(d) Show that the circumscribed and inscribed circles of △ ABC have the same center
...
Let △ ABC be a right triangle with C = 90◦
...
3 Identities
3
...
For example, we know
the reciprocal relations:
1
...
sec θ =
1
cos θ
when cos θ = 0
3
...
sin θ =
1
csc θ
when csc θ is defined and not 0
5
...
tan θ =
1
cot θ
when cot θ is defined and not 0
Notice that each of these equations is true for all angles θ for which both sides of the equation are defined
...
e
...
These
identities are often used to simplify complicated expressions or equations
...
1)
To prove this identity, pick a point ( x, y) on the terminal side of θ a distance r > 0 from the
origin, and suppose that cos θ = 0
...
r
65
66
Chapter 3 • Identities
§3
...
This is probably the most common technique
for proving identities
...
2)
y
We will now derive one of the most important trigonometric
identities
...
By the Pythagorean Theorem, r 2 = x2 + y2 (and hence r = x2 + y2 )
...
1
...
The same
argument holds if θ is in the other quadrants or on either axis
...
1
...
= sin θ , we can rewrite this as:
cos2 θ + sin2 θ = 1
(3
...
Note
that we use the notation sin2 θ to mean (sin θ )2 , likewise for cosine and the other trigonometric functions
...
From the above identity we can derive more identities
...
4)
cos2 θ = 1 − sin2 θ
(3
...
6)
cos θ = ±
1 − sin2 θ
(3
...
1
67
Also, from the inequalities 0 ≤ sin2 θ = 1 − cos2 θ ≤ 1 and 0 ≤ cos2 θ = 1 − sin2 θ ≤ 1, taking
square roots gives us the following bounds on sine and cosine:
− 1 ≤ sin θ ≤ 1
(3
...
9)
The above inequalities are not identities (since they are not equations), but they provide
useful checks on calculations
...
4
...
3), dividing both sides of the identity by cos2 θ gives
cos2 θ
sin2 θ
1
+
=
,
2 θ
2 θ
cos
cos
cos2 θ
so since tan θ =
sin θ
cos θ
and sec θ =
1
cos θ ,
we get:
1 + tan2 θ = sec2 θ
(3
...
3) by sin2 θ gives
cos2 θ
2
sin θ
so since cot θ =
cos θ
sin θ
and csc θ =
1
sin θ ,
+
sin2 θ
2
sin θ
=
1
sin2 θ
,
we get:
cot2 θ + 1 = csc2 θ
Example 3
...
Solution: We can use formula (3
...
2
Simplify 5 sin2 θ + 4 cos2 θ
...
5) to simplify:
5 sin2 θ + 4 cos2 θ = 5 sin2 θ + 4 1 − sin2 θ
= 5 sin2 θ + 4 − 4 sin2 θ
= sin2 θ + 4
(3
...
1
Example 3
...
Solution: We will expand the left side and show that it equals the right side:
cos θ
sin θ
+
cos θ
sin θ
(by (3
...
2))
=
sin θ sin θ
cos θ cos θ
·
+
·
cos θ sin θ
sin θ cos θ
(multiply both fractions by 1)
=
sin2 θ + cos2 θ
cos θ sin θ
(after getting a common denominator)
=
1
cos θ sin θ
(by (3
...
The reason is that, by its complexity, there will be
more things that you can do with that expression
...
Example 3
...
sec θ
Solution: Of the two sides, the left side looks more complicated, so we will expand that:
Prove that
1 + cot2 θ
csc2 θ
=
sec θ
sec θ
=
csc θ ·
(by (3
...
2))
Basic Trigonometric Identities • Section 3
...
5
tan2 θ + 2
= 1 + cos2 θ
...
10))
= 1 + cos2 θ
When trying to prove an identity where at least one side is a ratio of expressions, crossmultiplying can be an effective technique:
a
c
=
if and only if ad = bc
b
d
Example 3
...
cos θ
1 − sin θ
Solution: Cross-multiply and reduce both sides until it is clear that they are equal:
Prove that
(1 + sin θ )(1 − sin θ ) = cos θ · cos θ
1 − sin2 θ = cos2 θ
By (3
...
Thus, the original identity holds
...
7
Suppose that a cos θ = b and c sin θ = d for some angle θ and some constants a, b, c, and d
...
Solution: Multiply both sides of the first equation by c and the second equation by a:
ac cos θ = bc
ac sin θ = ad
Now square each of the above equations then add them together to get:
(ac cos θ )2 + (ac sin θ )2 = ( bc)2 + (ad )2
(ac)2 cos2 θ + sin2 θ = b2 c2 + a2 d 2
a2 c 2 = b 2 c 2 + a2 d 2
(by (3
...
The trick was to get a common coefficient (ac) for
cos θ and sin θ so that we could use cos2 θ + sin2 θ = 1
...
70
Chapter 3 • Identities
§3
...
We showed that sin θ = ± 1 − cos2 θ for all θ
...
2
...
1 − sin2 θ for all θ
...
Suppose that you are given a system of two equations of the following form:1
A cos φ = B ν1 − Bν2 cos θ
A sin φ = B ν2 sin θ
...
1
2
For Exercises 4-16, prove the given identity
...
cos θ tan θ = sin θ
5
...
7
...
= 1 − sin θ
1 + sin θ
10
...
14
...
sin θ = ±
= tan2 θ
tan θ
1 + tan2 θ
csc θ
= csc2 θ
sin θ
1 − 2 cos2 θ
9
...
cos4 θ − sin4 θ = 1 − 2 sin2 θ
13
...
tan θ + tan φ
= tan θ tan φ
cot θ + cot φ
1 − tan2 θ
1 − cot2 θ
= 1 − sec2 θ
(Hint: Solve for sin2 θ in Exercise 14
...
Sometimes identities can be proved by geometrical methods
...
1
...
What
must y equal? Use that to prove the identity for acute θ
...
1
...
Does the identity hold if θ is on either
axis?
(1, y)
y
y
θ
x
0
1
Figure 3
...
2
18
...
19
...
20
...
Use r 2 = x2 + y2 to show that r = a (1 − ǫ cos ψ)
...
)
21
...
1 These types of equations arise in physics, e
...
in the study of photon-electron collisions
...
95-97 in W
...
, 1960
...
2
71
3
...
For the sum of any two angles A and B, we have the addition formulas:
sin ( A + B) = sin A cos B + cos A sin B
(3
...
13)
To prove these, first assume that A and B are acute angles
...
2
...
Note in both cases that ∠ QPR = A , since
∠ QPR = ∠ QPO − ∠ OP M = (90◦ − B) − (90◦ − ( A + B)) = A in Figure 3
...
1(a), and
∠ QPR = ∠ QPO + ∠ OP M = (90◦ − B) + (90◦ − (180◦ − ( A + B))) = A in Figure 3
...
1(b)
...
2
...
14)
72
Chapter 3 • Identities
§3
...
(3
...
It is simple to verify that they
hold in the special case of A = B = 0◦
...
5 which involve adding or subtracting 90◦ :
sin (θ + 90◦ ) =
cos θ
cos (θ + 90◦ ) = − sin θ
sin (θ − 90◦ ) = − cos θ
cos (θ − 90◦ ) =
sin θ
These will be useful because any angle can be written as the sum of an acute angle (or
0◦ ) and integer multiples of ±90◦
...
So if we can prove that the identities hold when adding or subtracting 90◦
to or from either A or B, respectively, where A and B are acute or 0◦ , then the identities
will also hold when repeatedly adding or subtracting 90◦ , and hence will hold for all angles
...
15))
= sin ( A + 90◦ ) cos B + cos ( A + 90◦ ) sin B ,
so the identity holds for A + 90◦ and B (and, similarly, for A and B + 90◦ )
...
Thus, the addition
formula (3
...
A similar argument shows that the addition
formula (3
...
QED
Replacing B by −B in the addition formulas and using the relations sin (−θ ) = − sin θ and
cos (−θ ) = cos θ from Section 1
...
16)
cos ( A − B) = cos A cos B + sin A sin B
(3
...
2
73
sin
Using the identity tan θ = cos θ , and the addition formulas for sine and cosine, we can
θ
derive the addition formula for tangent:
tan ( A + B) =
=
sin ( A + B)
cos ( A + B)
sin A cos B + cos A sin B
cos A cos B − sin A sin B
sin
= cos
cos
cos
A
A
A
A
cos B
cos A sin B
+
cos B
cos A cos B
cos B
sin A sin B
−
cos B
cos A cos B
(divide top and bottom by cos A cos B)
B
cos sin B
A
sin A cos
· + ·
tan A + tan B
cos
cos
A cos B =
= cos A B
sin A sin B
1 − tan A tan B
·
1 −
cos A cos B
This, combined with replacing B by −B and using the relation tan (−θ ) = − tan θ , gives us
the addition and subtraction formulas for tangent:
tan ( A + B) =
tan A + tan B
1 − tan A tan B
(3
...
19)
Example 3
...
12
13 ,
and cos B =
5
13 ,
Solution: Using the addition formula for sine, we get:
sin ( A + B) = sin A cos B + cos A sin B
=
4 5
3 12
·
+
·
5 13
5 13
sin ( A + B) =
⇒
56
65
Using the addition formula for cosine, we get:
cos ( A + B) = cos A cos B − sin A sin B
=
4 12
3 5
·
−
·
5 13
5 13
cos ( A + B) = −
⇒
33
65
Instead of using the addition formula for tangent, we can use the results above:
56
tan ( A + B) =
sin ( A + B)
= 65
33
cos ( A + B)
− 65
⇒
tan ( A + B) = −
56
33
find the exact values
74
Chapter 3 • Identities
§3
...
9
Prove the following identity:
sin ( A + B + C ) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C
Solution: Treat A + B + C as ( A + B) + C and use the addition formulas three times:
sin ( A + B + C ) = sin (( A + B) + C )
= sin ( A + B) cos C + cos ( A + B) sin C
= (sin A cos B + cos A sin B) cos C + (cos A cos B − sin A sin B) sin C
Example 3
...
Solution: Note that this is not an identity which holds for all angles; since A , B, and C are the
angles of a triangle, it holds when A , B, C > 0◦ and A + B + C = 180◦
...
5, we get:
tan A + tan B + tan C = tan A + tan B + tan (180◦ − ( A + B))
= tan A + tan B − tan ( A + B)
tan A + tan B
1 − tan A tan B
1
(tan A + tan B) 1 −
1 − tan A tan B
1 − tan A tan B
1
(tan A + tan B)
−
1 − tan A tan B
1 − tan A tan B
− tan A tan B
(tan A + tan B) ·
1 − tan A tan B
tan A + tan B
tan A tan B · −
1 − tan A tan B
= tan A + tan B −
=
=
=
=
= tan A tan B · (− tan ( A + B))
= tan A tan B · (tan (180◦ − ( A + B)))
Example 3
...
Show that2
sin A sin B + sin C sin D = sin ( A + C ) sin (B + C )
...
However,
notice that the right side has no D term
...
2 This is the “trigonometric form” of Ptolemy’s Theorem, which says that a quadrilateral can be inscribed in a
circle if and only if the sum of the products of its opposite sides equals the product of its diagonals
...
2
75
So since sin D = sin ( A + B + C ), we get
sin A sin B + sin C sin D = sin A sin B + sin C sin ( A + B + C ) , so by Example 3
...
It may not be immediately obvious where to go from here, but it is not completely guesswork
...
There are two terms involving cos B sin C , so group them together to get
sin A sin B + sin C sin D = sin A sin B − sin C sin A sin B sin C + sin C cos A sin B cos C
+ cos B sin C (sin A cos C + cos A sin C )
= sin A sin B (1 − sin2 C ) + sin C cos A sin B cos C
+ cos B sin C sin ( A + C )
= sin A sin B cos2 C + sin C cos A sin B cos C
+ cos B sin C sin ( A + C )
...
12
In the study of the propagation of electromagnetic waves, Snell’s law gives the relation
n 1 sin θ1 = n 2 sin θ2 ,
(3
...
The quantity
r1 2 s =
n 1 cos θ1 − n 2 cos θ2
n 1 cos θ1 + n 2 cos θ2
(3
...
Show
that this can be written as:
sin (θ2 − θ1 )
r1 2 s =
sin (θ2 + θ1 )
76
Chapter 3 • Identities
§3
...
This is a common technique
...
Verify the addition formulas (3
...
13) for A = B = 0◦
...
2
...
sin A =
40
41 ,
cos A =
9
41 ,
sin B =
20
29 ,
cos B =
21
29
4
...
5
...
6
...
Explain why this shows that
− 2 ≤ sin θ + cos θ ≤
2
for all angles θ
...
7
...
tan ( A + B + C ) =
9
...
cot ( A − B) =
cot A cot B + 1
cot B − cot A
Sum and Difference Formulas • Section 3
...
cot A + cot B =
sin A sin B
77
cos ( A + B)
= cot A − tan B
sin A cos B
sin ( A − B)
cot B − cot A
14
...
tan (θ + 45◦ ) =
12
...
Generalize Exercise 6: For any a and b, − a2 + b2 ≤ a sin θ + b cos θ ≤
a2 + b2 for all θ
...
Continuing Example 3
...
22)
t1 2 s =
n 1 cos θ1 + n 2 cos θ2
can be written as:
t1 2 s =
2 cos θ1 sin θ2
sin (θ2 + θ1 )
17
...
e
...
Show that
tan θ =
y
y = m2 x + b2
θ
m1 − m2
...
26 from Section 1
...
)
18
...
19
...
(Hint: Use Exercise 9 and C = 180◦ − ( A + B)
...
For any positive angles A , B, and C such that A + B + C = 90◦ , show that
tan A tan B + tan B tan C + tan C tan A = 1
...
Prove the identity sin ( A + B) cos B − cos ( A + B) sin B = sin A
...
22
...
What are the endpoint’s
new coordinates ( x′ , y′ ) after a counterclockwise rotation by an angle β ? Your answer should be in
terms of r , α, and β
...
3
3
...
23)
cos 2θ = cos θ − sin θ
(3
...
25)
2
2
2 tan θ
1 − tan2 θ
To derive the sine double-angle formula, we see that
sin 2θ = sin (θ + θ ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ
...
26)
(3
...
13
Prove that sin 3θ = 3 sin θ − 4 sin3 θ
...
23) and
(3
...
3
79
Example 3
...
(1 + tan2 z)2
Solution: Expand the right side and use 1 + tan2 z = sec2 z :
Prove that sin 4 z =
2
4 tan z (1 − tan z)
(1 + tan2 z)2
=
=
4·
4·
cos2 z sin2 z
sin z
−
·
cos z
cos2 z cos2 z
(sec2 z)2
sin z cos 2 z
·
cos z cos2 z
2
1
2 z
cos
(by formula (3
...
23))
(by formula (3
...
3
Closely related to the double-angle formulas are the half-angle formulas:
1 − cos θ
2
1 + cos θ
2 1
cos 2 θ =
2
1 − cos θ
1
tan2 2 θ =
1 + cos θ
sin2 1 θ =
2
(3
...
29)
(3
...
g
...
3 See p
...
A
...
I
...
80
Chapter 3 • Identities
§3
...
31)
1
cos 2 θ = ±
1 + cos θ
2
(3
...
33)
In the above form, the sign in front of the square root is determined by the quadrant in
1
which the angle 2 θ is located
...
So in this
2
1
case cos 1 θ < 0 and hence we would have cos 2 θ = − 1 + cos θ
...
33), multiplying the numerator and denominator inside the square root by
(1 − cos θ ) gives
tan 1 θ = ±
2
1 − cos θ 1 − cos θ
·
= ±
1 + cos θ 1 − cos θ
(1 − cos θ )2
= ±
1 − cos2 θ
(1 − cos θ )2
2
sin θ
= ±
1 − cos θ
...
Thus, the minus sign in front of the last expression is not possible (since that
1
would switch the signs of tan 2 θ and sin θ ), so we have:
1
tan 2 θ =
1 − cos θ
sin θ
(3
...
34) by 1 + cos θ gives
1
tan 2 θ =
1 − cos θ 1 + cos θ
1 − cos2 θ
sin2 θ
·
=
=
,
sin θ
1 + cos θ
sin θ (1 + cos θ )
sin θ (1 + cos θ )
so we also get:
1
tan 2 θ =
sin θ
1 + cos θ
(3
...
34) and (3
...
36)
Double-Angle and Half-Angle Formulas • Section 3
...
15
2 sec θ
...
29) for cos2
gives us
2
sec θ
2 sec θ
2
1
=
·
=
...
1
2
...
cos 3θ = 4 cos3 θ − 3 cos θ
sin 2θ
cos 2θ
−
= sec θ
sin θ
cos θ
2
5
...
tan2 1 θ =
2
tan θ + sin θ
sin 3θ
cos 3θ
−
= 2
sin θ
cos θ
3 tan θ − tan3 θ
6
...
=
1 + cos 2θ
cos2 θ
3
...
9
...
Give an example of a specific angle θ that would make that equation false
...
10
...
θ
◦
0 − 90◦
90◦ − 180◦
180◦ − 270◦
270◦ − 360◦
1
2θ
0◦ − 45◦
45◦ − 90◦
90◦ − 135◦
135◦ − 180◦
sin θ
+
1
tan 2 θ
+
θ
◦
360 − 450◦
450◦ − 540◦
540◦ − 630◦
630◦ − 720◦
1
2θ
sin θ
180◦ − 225◦
225◦ − 270◦
270◦ − 315◦
315◦ − 360◦
11
...
For Exercises 12-17, prove the given identity for any right triangle △ ABC with C = 90◦
...
sin ( A − B) = cos 2B
2 ab
14
...
tan 2 A = 2
b − a2
13
...
tan 2 A =
a
c+b
15
...
Continuing Exercise 20 from Section 3
...
Show that tan 1 θ =
2
1+ǫ
1−ǫ
tan 1 ψ
...
4
3
...
It is very common
to encounter terms such as sin A + sin B or sin A cos B in calculations, so we will now
derive identities for those expressions
...
37)
(sin ( A + B) − sin ( A − B))
(3
...
39)
(cos ( A + B) − cos ( A − B))
(3
...
We
see that
✭
✭
✭
✭
sin ( A + B) + sin ( A − B) = (sin A cos B + ✭✭A ✭ B) + (sin A cos B − ✭✭A ✭ B)
cos ✭ sin
cos ✭ sin
= 2 sin A cos B ,
so formula (3
...
Notice how in each of the above
identities a product (e
...
sin A cos B) of trigonometric functions is shown to be equivalent to
1
a sum (e
...
2 (sin ( A + B) + sin ( A − B))) of such functions
...
41)
sin A − sin B =
1
1
2 cos 2 ( A + B) sin 2 ( A − B)
(3
...
43)
cos A − cos B = −2 sin
1
2 ( A + B)
1
2 ( A + B)
cos
sin
1
2 ( A − B)
1
2 ( A − B)
(3
...
Then x + y = A and x − y = B
...
43), make the above substitutions in formula (3
...
39))
cos
1
2 ( A − B)
...
Other Identities • Section 3
...
16
We are now in a position to prove Mollweide’s equations, which we introduced in Section 2
...
1
c
c
cos 1 C
sin 2 C
2
First, since C = 2 ·
1
2 C,
1
by the double-angle formula we have sin C = 2 sin 1 C cos 2 C
...
42))
(since A + B = 180◦ − C )
(since cos (90◦ − 1 C ) = sin 1 C )
...
The proof of the other equation is similar (see Exercise 7)
...
17
Using Mollweide’s equations, we can prove the Law of Tangents: For any triangle △ ABC ,
1
tan 2 ( A − B)
a−b
,
=
1
a+b
tan 2 ( A + B)
1
tan 2 (B − C )
b−c
,
=
b+c
tan 1 (B + C )
2
1
tan 2 (C − A )
c−a
...
We
see that
sin 1 ( A − B)
2
a−b
cos 1 C
a−b
c
2
=
=
a+b
a+b
cos 1 ( A − B)
2
c
sin 1 C
2
=
1
sin 2 ( A − B)
cos 1 ( A − B)
2
·
(by Mollweide’s equations)
1
sin 2 C
cos 1 C
2
1
1
= tan 1 ( A − B) · tan 2 C = tan 1 ( A − B) · tan (90◦ − 2 ( A + B))
2
2
1
1
= tan 2 ( A − B) · cot 2 ( A + B)
=
tan 1 ( A − B)
2
tan 1 ( A + B)
2
...
5)
2
84
Chapter 3 • Identities
§3
...
18
For any triangle △ ABC , show that
1
2A
cos A + cos B + cos C = 1 + 4 sin
1
sin 2 B sin 1 C
...
43)
1
1
= 1 + 2 cos 2 ( A + B + 2C ) cos 1 ( A + B) + 2 cos 1 ( A + B) cos 2 ( A − B)
2
2
1
1
= 1 + 2 cos 2 ( A + B) cos 2 ( A + B + 2C ) + cos 1 ( A − B)
2
, so
1
= 1 + 2 cos 2 ( A + B) · 2 cos 1 ( A + C ) cos 1 (B + C ) by formula (3
...
Thus,
2
1
1
cos A + cos B + cos C = 1 + 4 cos (90◦ − 2 C ) cos (90◦ − 1 B) cos (90◦ − 2 A )
2
1
= 1 + 4 sin 1 C sin 2 B sin
2
= 1 + 4 sin
1
2A
sin
1
2B
sin
1
2A
1
2C
, so rearranging the order gives
...
19
For any triangle △ ABC , show that sin
Solution: Let u = sin
1
2A
sin
1
2B
1
2A
1
2 C
...
Apply formula (3
...
18
...
Notice that the expression above is a quadratic equation in the term
cos 1 ( A + B)
...
But we know
that cos 1 ( A + B) is a real number (and, hence, a solution exists), so we must have
2
cos
cos2
1
2 ( A − B)
− 8u ≥ 0
⇒
=
u ≤
1
8
cos2
1
2 ( A − B)
≤
1
8
sin
⇒
1
2A
1
sin 2 B sin 1 C ≤
2
Example 3
...
are all positive, so
sin 1 B sin 1 C > 1
2
2
by Example 3
...
Also, by Examples 3
...
19 we have
cos A + cos B + cos C = 1 + 4 sin
Hence, 1 < cos A + cos B + cos C ≤
3
2
...
1
8
...
4
85
Example 3
...
12 in Section 3
...
Use it to show that the
p-polarization transmission Fresnel coefficient defined by
t1 2 p =
2 n 1 cos θ1
n 2 cos θ1 + n 1 cos θ2
(3
...
sin (θ1 + θ2 ) cos (θ1 − θ2 )
Solution: Multiply the top and bottom of t 1 2 p by sin θ1 sin θ2 to get:
t1 2 p =
t1 2 p =
sin θ1 sin θ2
2 n 1 cos θ1
·
n 2 cos θ1 + n 1 cos θ2 sin θ1 sin θ2
=
2 ( n 1 sin θ1 ) cos θ1 sin θ2
( n 2 sin θ2 ) sin θ1 cos θ1 + ( n 1 sin θ1 ) sin θ2 cos θ2
=
2 cos θ1 sin θ2
sin θ1 cos θ1 + sin θ2 cos θ2
=
=
=
2 cos θ1 sin θ2
1
2
1
2
(by the double-angle formula)
(sin 2 θ1 + sin 2θ2 )
(2 sin
(by Snell’s law)
2 cos θ1 sin θ2
1
1
2 (2θ1 + 2θ2 ) cos 2 (2θ1 − 2θ2 ))
(by formula (3
...
22
In an AC electrical circuit, the instantaneous power p( t) delivered to the entire circuit in the sinusoidal steady state at time t is given by
p ( t) = v( t) i ( t) ,
where the voltage v( t) and current i ( t) are given by
v( t) = Vm cos ω t ,
i ( t) = I m cos (ω t + φ) ,
for some constants Vm , I m , ω, and φ
...
Solution: By definition of p( t), we have
p( t) = Vm I m cos ω t cos (ω t + φ)
1
2 (cos (2ω t + φ) +
1
1
2 Vm I m cos φ + 2 Vm I m
= Vm I m ·
cos (−φ))
(by formula (3
...
86
Chapter 3 • Identities
§3
...
Prove formula (3
...
2
...
39)
...
Prove formula (3
...
4
...
41)
...
Prove formula (3
...
6
...
44)
...
Prove Mollweide’s second equation: For any triangle △ ABC ,
1
cos 2 ( A − B)
a+b
...
Continuing Example 3
...
46)
r1 2 p =
n 2 cos θ1 + n 1 cos θ2
can be written as:
r1 2 p =
tan (θ1 − θ2 )
tan (θ1 + θ2 )
9
...
22
...
4 Show that p( t) can be written as
p ( t) =
1
2
Vm I m cos (θ − φ) +
1
2
Vm I m cos (2ω t + θ + φ)
...
1
1
10
...
18 using (sin A + sin B) + (sin C − sin ( A + B + C ))
...
cos A + cos (B − C ) = 2 sin B sin C
12
...
)
13
...
)
14
...
1
2A
=
1
(sin A + sin B) ≤ sin 2 ( A + B)
1
1
(Hint: Show that sin 2 ( A + B) − 2 (sin A + sin B) ≥ 0
...
In Example 3
...
We
will discuss their unit of measurement in Chapter 4
...
1 Radians and Degrees
So far we have been using degrees as our unit of measurement for angles
...
The idea is simple: associate
a central angle of a circle with the arc that it intercepts
...
1
...
In geometry you learned that the
circumference C of the circle is C = 2 π r , where π = 3
...
1
AB = 4 C =
π
2
1
AB = 2 C = π r
B
90◦
O
A
B
(a) θ = 90◦
AB = C = 2 π r
180◦
r
360◦
A
O
O
(b) θ = 180◦
A
B
(c) θ = 360◦
Figure 4
...
1 Angle θ and intercepted arc AB on circle of circumference C = 2π r
In Figure 4
...
1 we see that a central angle of 90◦ cuts off an arc of length π r , a central
2
angle of 180◦ cuts off an arc of length π r , and a central angle of 360◦ cuts off an arc of length
2π r , which is the same as the circumference of the circle
...
The radius r was arbitrary, but the 2π in front of it stays the same
...
1)
The above relation gives us any easy way to convert between degrees and radians:
Degrees to radians:
x degrees
Radians to degrees:
x radians =
=
π
·x
180
180
·x
π
radians
(4
...
3)
87
88
Chapter 4 • Radian Measure
§4
...
2) follows by dividing both sides of equation (4
...
Formula (4
...
1) by 2π then multiplying both sides by x
...
When an angle is given as some multiple of π, you can
assume that the units being used are radians
...
1
Convert 18◦ to radians
...
2) for degrees to radians, we get
18◦ =
π
π
· 18 =
rad
...
2
Convert
π
9
radians to degrees
...
3) for radians to degrees, we get
π
9
rad =
180 π
·
= 20◦
...
1 Commonly used angles in radians
Degrees
0◦
30◦
45◦
60◦
Radians
0
π
6
π
4
π
3
Degrees
90◦
Radians
π
2
Degrees
Radians
Degrees
Radians
180◦
π
270◦
3π
2
120◦
2π
3
210◦
7π
6
300◦
5π
3
135◦
3π
4
225◦
5π
4
315◦
7π
4
150◦
5π
6
240◦
4π
3
330◦
11π
6
Table 4
...
Using the conversion formula (4
...
3◦
...
1
...
This definition
does not depend on the choice of r (imagine resizing Figure 4
...
2)
...
1
...
1
89
One reason why radians are used is that the scale is smaller than for degrees
...
283185307, which is much smaller than 360, the number of degrees
in one revolution
...
Another
reason is that often in physical applications the variables being used are in terms of arc
length, which makes radians a natural choice
...
On
✄
many calculators there is a button labeled ✂DRG ✁ for switching between degree mode (D),
radian mode (R), and gradian mode (G)
...
Make sure that your
calculator is in the correct angle mode before entering angles, or your answers will likely be
way off
...
0698 ,
sin (4 rad) = −0
...
Using your
in
✄
✄
sin , cos ,
tan
calculator’s ✂ −1 ✁ ✂ −1 ✁ and ✂ −1 ✁buttons in radian mode will of course give you the angle
as a decimal, not an expression in terms of π
...
2
Exercises
For Exercises 1-5, convert the given angle to radians
...
4◦
2
...
130◦
For Exercises 6-10, convert the given angle to degrees
...
4 rad
rad
7
...
rad
5
9
4
...
29π
rad
30
5
...
35 rad
11
...
Whatever the answer is, take its
cosine
...
Keep repeating this
...
What is that number?
Try starting with a number different from 0
...
Does the same thing happen? If so, does it take fewer iterations for the answer to start repeating
than in radian mode, or more?
1 A gradian is defined as 1 of a circle, i
...
there are 400 gradians in one revolution
...
Outside of a few specialized areas (e
...
artillery calculations), gradians are nevertheless not widely used today
...
90
Chapter 4 • Radian Measure
§4
...
2 Arc Length
In Section 4
...
Note that 2π is
the ratio of the circumference (i
...
total arc length) C of a circle to its radius r :
Radian measure of 1 revolution = 2π =
2π r
C
total arc length
=
=
r
r
radius
Clearly, that ratio is independent of r
...
To see this, recall our formal definition of a radian: the central angle in a circle of radius r
which intercepts an arc of length r
...
2
...
Clearly, the combined central angle of the two angles has radian
measure 1 + 1 = 2, and the combined arc length is r + r = 2 r
...
2
...
2
...
Clearly, this cuts the arc length r in half as well
...
radius
Arc Length • Section 4
...
The above discussion says more, namely that the
ratio of the length s of an intercepted arc to the radius r is preserved, precisely because that
ratio is the measure of the central angle in radians (see Figure 4
...
2)
...
2
...
Then the arc length s is:
s = rθ
(4
...
3
In a circle of radius r = 2 cm, what is the length s of the arc intercepted by a central angle of measure
θ = 1
...
4), we get:
s = r θ = (2) (1
...
4 cm
Example 4
...
4) blindly with θ = 41◦ , we would get s = r θ = (10) (41) = 410 ft
...
83 ft! Our error was
in using the angle θ measured in degrees, not radians
...
716 rad ⇒ s = r θ = (10) (0
...
16 ft
θ = 41◦ =
180
Note that since the arc length s and radius r are usually given in the same units, radian
s
measure is really unitless, since you can think of the units canceling in the ratio r , which is
just θ
...
92
Chapter 4 • Radian Measure
§4
...
5
A central angle in a circle of radius 5 m cuts off an arc of length 2 m
...
4), we get:
θ =
2
s
=
= 0
...
4 rad =
180
· 0
...
92◦
π
For central angles θ > 2π rad, i
...
θ > 360◦ , it may not be clear what is meant by the intercepted arc, since the angle is larger than one revolution and hence “wraps around” the circle
more than once
...
In other words, formula (4
...
What about negative angles? In this case using s = r θ would mean that the arc length is
negative, which violates the usual concept of length
...
Example 4
...
A
cylindrical container with a radius of 2 ft is pushed away from the wall as
far as it can go while being held in by the rope, as in Figure 4
...
3 which
shows the top view
...
Also, notice that △ ADE is a right triangle, so the hypotenuse
has length AE = DE 2 + D A 2 = 32 + 42 = 5 ft, by the Pythagorean Theorem
...
So by the Pythagorean Theorem we have
AB =
AE 2 − BE 2 =
52 − 22 =
21 ft
...
2
...
4) the arc BC has length BE · θ , where θ = ∠ BEC is the
supplement of ∠ AED + ∠ AEB
...
1◦
3
and
cos ∠ AEB =
BE
2
=
⇒ ∠ AEB = 66
...
1◦ + 66
...
5◦
...
5 = 1
...
Thus,
L = 2 ( AB + ·BC ) = 2 ( 21 + BE · θ ) = 2 ( 21 + (2) (1
...
4 ft
...
2
93
Example 4
...
Find the
total length L of the belt around the pulleys
...
2
...
F
E
5
C
5
D
3
B
A
G
15
Figure 4
...
4 Belt pulleys with radii 5 cm and 8 cm
First, at the center B of the pulley with radius 8, draw a circle of radius 3, which is the difference
in the radii of the two pulleys
...
Then we know that
the tangent line AC to this smaller circle is perpendicular to the line segment BF
...
Now since AE ⊥ EF and EF ⊥ CF and CF ⊥ AC , the quadrilateral
AEFC must be a rectangle
...
By formula (4
...
So thinking of angles in radians (using π rad = 180◦ ), we see from Figure 4
...
4
that
π
π
− ∠ BAC ,
∠ D AE = π − ∠ E AC − ∠ BAC = π − − ∠ BAC =
2
2
where
3
BC
=
= 0
...
201 rad
...
201 = 1
...
So since AE and BF are parallel, we have ∠ ABC = ∠ D AE =
2
1
...
Thus, ∠ GBF = π − ∠ ABC = π − 1
...
77 rad
...
37) + 6 6 + 8 (1
...
41 cm
...
2
Exercises
For Exercises 1-4, find the length of the arc cut off by the given central angle θ in a circle of radius r
...
θ = 171◦ , r = 8 m
1
...
8 rad, r = 12 cm
3
...
A central angle in a circle of radius 2 cm cuts off an arc of length 4
...
What is the measure of
the angle in radians? What is the measure of the angle in degrees?
5
...
Find the total length L of the belt around the pulleys
...
In Figure 4
...
5 one end of a 4 ft iron rod is attached to the center of a pulley with radius 0
...
The other end is attached at a 40◦ angle to a wall, at a spot 6 ft above the lower end of a steel wire
supporting a box
...
Find the length L of the wire from the wall to the box
...
2
...
2
...
Figure 4
...
6 shows the same setup as in Exercise 6 but now the wire comes out of the wall 2 ft
above where the rod is attached
...
8
...
2
...
9
...
(Hint: Draw a
circle of radius 5 centered at A , then draw a
tangent line to that circle from B
...
2
...
Suppose that in Figure 4
...
7 the lines do not criss-cross but instead go straight across, as in a
belt pulley system
...
11
...
12
...
e
...
Compare it to the circumference of the circle
...
3
95
4
...
We
will now learn how to find the area of a sector of a circle
...
3
...
Let θ be a central angle in a circle of radius r and let A be the area of its
sector
...
In other words, again using
radian measure,
area of sector
sector angle
=
area of entire circle
one revolution
⇒
r
θ
Figure 4
...
1
θ
A
=
...
5)
where θ is measured in radians
...
8
Find the area of a sector whose angle is
Solution: Using θ =
π
5
π
5
rad in a circle of radius 4 cm
...
5), the area A of the sector is
A =
1 2
2r
θ =
1
2
(4)2 ·
π
5
=
8π
5
cm2
...
9
Find the area of a sector whose angle is 117◦ in a circle of radius 3
...
Solution: As with arc length, we have to make sure that the angle is measured in radians or else
the answer will be way off
...
5 in formula (4
...
042 rad
⇒
A =
1 2
2r
θ =
1
2
(3
...
042) = 12
...
For a sector whose angle is θ in a circle of radius r , the length of the arc cut off by that
angle is s = r θ
...
5) the area A of the sector can be written as:3
A =
1
2
rs
(4
...
3 In some texts this formula is taken as a result from elementary geometry and then used to prove formula
(4
...
96
Chapter 4 • Radian Measure
§4
...
10
Find the area of a sector whose arc is 6 cm in a circle of radius 9 cm
...
6) for the area A , we get
A =
1
2
rs = =
Note that the angle subtended by the arc is θ =
1
2
(9) (6) = 27 cm2
...
Example 4
...
7 in Section 4
...
Solution: Recall that the belt pulleys have radii of 5 cm and 8 cm, and their centers are 15 cm apart
...
7 that EF = AC = 6 6, ∠ D AE = 1
...
77 rad
...
3
...
F
6 6
E
5
6 6
5
1
...
37
D
C
B
A
G
15
Figure 4
...
2 Belt pulleys with radii 5 cm and 8 cm
Since AEFC is a rectangle with sides 5 and 6 6, its area is 30 6
...
Thus, using formula (4
...
37) + 30 6 + 9 6 +
= 338
...
1
2
(8)2 (1
...
3
97
So far we have dealt with the area cut off by a central angle
...
3
...
Drawing line segments
b
from the center of the circle to the endpoints of the chords indicates
how to solve this problem: add up the areas of the two triangles and
Figure 4
...
3
the sector formed by the central angle
...
Also, recall (Theorem 2
...
5) that a central angle has twice the measure of any
inscribed angle which intercepts the same arc
...
Another type of region we can consider is a segment of a circle,
which is the region between a chord and the arc it cuts off
...
3
...
By formula (2
...
4 for the area of a triangle given two sides and their
included angle, we know that
area of △ O AB =
1
2 ( r ) ( r ) sin
θ =
1 2
2r
1 2
2r
θ −
r
θ
B
r
O
sin θ
...
3
...
(4
...
7) we must have θ > sin θ for 0 < θ ≤ π (measured
in radians), since the area of a segment is positive for those angles
...
12
Find the area of the segment formed by a chord of length 3 in a circle of
radius 2
...
3
...
We can use the Law of Cosines to find the subtended
central angle θ :
cos θ =
22 + 22 − 32
= −0
...
696 rad
Thus, by formula (4
...
696 − sin 1
...
408
Figure 4
...
5
98
Chapter 4 • Radian Measure
§4
...
13
The centers of two circles are 7 cm apart, with one circle having a radius of 5 cm and the other a
radius of 3 cm
...
Solution: In Figure 4
...
6(a), we see that the intersection of the two circles is the union of the segments formed by the chord CD in each circle
...
C
5
A
4
B
7
C
5
D
4
α
A
β
7
1
(a) ∠ BAC = 2 ∠ C AD , ∠ ABC = 1 ∠ CBD
2
B
(b) Triangle △ ABC
Figure 4
...
6
1
By symmetry, we see that ∠ BAC = 2 ∠ C AD and ∠ ABC =
∠ ABC , as in Figure 4
...
6(b)
...
8286
2 (7) (5)
72 + 42 − 52
= 0
...
So let α = ∠ BAC and β =
⇒
α = 0
...
594) = 1
...
775 rad
⇒
∠ CBD = 2 (0
...
550 rad
Thus, the area K is
K = (Area of segment CD in circle at A ) + (Area of segment CD in circle at B)
=
1
2
(5)2 (1
...
188) +
1
2
(4)2 (1
...
550)
= 7
...
Exercises
For Exercises 1-3, find the area of the sector for the given angle θ and radius r
...
θ = 2
...
2 cm
2
...
5 ft
3
...
The centers of two belt pulleys, with radii of 3 cm and 6 cm, respectively, are 13 cm apart
...
5
...
Find the total area K
enclosed by the belt
...
3
99
6
...
2
...
7
...
r = a, s = a
9
...
10
...
a = 1 cm, r = 5 cm
12
...
Find the area of the shaded region in Figure 4
...
7
...
3
...
3
...
3
...
Find the area of the shaded region in Figure 4
...
8
...
)
15
...
3
...
16
...
Find the area of their intersection
...
Three circles with radii of 4 m, 2 m, and 1 m are externally tangent to each other
...
3
...
(Hint: Connect the centers of the
circles
...
3
...
3
...
Show that the total area enclosed by the loop around the three circles of radius r in Figure 4
...
11
is (π + 6 + 3) r 2
...
For a fixed central angle θ , how much does the area of its sector increase when the radius of the
circle is doubled? How much does the length of its intercepted arc increase?
100
Chapter 4 • Radian Measure
§4
...
4 Circular Motion: Linear and Angular Speed
Radian measure and arc length can be applied to the study
of circular motion
...
4
...
Then it makes sense to define the (average) linear
speed ν of the object as:
ν =
time t > 0
s
t
Figure 4
...
1
(4
...
Then we define the
(average) angular speed ω of the object as:
ω =
θ
t
(4
...
Linear
speed is measured in distance units per unit time (e
...
feet per second)
...
We will usually omit the word average when discussing linear and angular speed here
...
10)
4 Many trigonometry texts assume uniform motion, i
...
constant speeds
...
Also,
many texts use the word velocity instead of speed
...
Circular Motion: Linear and Angular Speed • Section 4
...
14
An object sweeps out a central angle of π radians in 0
...
Find its linear and angular speed over that time period
...
5 sec, r = 3 m, and θ =
π
3
rad
...
5 sec
θ
ω =
⇒
2π
rad/sec ,
3
and thus the linear speed ν is
ν = ωr =
2π
rad/sec (3 m)
3
⇒
ν = 2π m/sec
...
Recall that radians are actually
unitless, which is why in the formula ν = ω r the radian units disappear
...
15
An object travels a distance of 35 ft in 2
...
Find its
linear and angular speed over that time period
...
7 sec, r = 2 ft, and s = 35 ft
...
7 sec
⇒
ν = 12
...
96 ft/sec = ω (2 ft)
⇒
ω = 6
...
Example 4
...
How large of a
central angle does it sweep out in 3
...
1 sec, ν = 10 m/sec, and r = 4 m
...
1 sec)
=
=
= 7
...
r
r
4m
In many physical applications angular speed is given in revolutions per minute, abbreviated as rpm
...
Converting to other units for angular
speed works in a similar way
...
4
Example 4
...
What is the angular speed ω1 of the larger gear?
Solution: Imagine a particle on the outer radius of each gear
...
In other
words, s 1 = s 2 , where s 1 and s 2 are the distances traveled by
the particles on the gears with radii r 1 and r 2 , respectively
...
Thus,
ν1 t = ν2 t
⇒
r 1 = 5 cm
ω2 = 25 rpm
r 2 = 4 cm
ν1 = ν2 ,
so by formula (4
...
4
...
11)
Note that this holds for any two gears
...
Exercises
For Exercises 1-6, assume that a particle moves along a circle of radius r for a period of time t
...
1
...
r = 8 m, t = 2 sec, θ = 3 rad
3
...
2 sec, θ = 172◦
4
...
6 sec, s = 3 m
5
...
6 sec, s = 6 m
6
...
5 ft, t = 0
...
An object moves at a constant linear speed of 6 m/sec around a circle of radius 3
...
How large
of a central angle does it sweep out in 1
...
Two interlocking gears have outer radii of 6 cm and 9 cm, respectively
...
Three interlocking gears have outer radii of 2 cm, 3 cm, and 4 cm, respectively
...
In Example 4
...
11) still hold if the radii r 1 and r 2 are replaced by the number
of teeth N1 and N2 , respectively, of the two gears as shown in Figure 4
...
2?
11
...
What is the linear speed of a speck of dust
on the outer edge of the record in inches per second?
12
...
Show that α = ω2 r , where ω is the angular speed
...
In the graphs we will always use radians for the angle measure
...
1 Graphing the Trigonometric Functions
y
The first function we will graph is the sine function
...
This is the circle of radius
1
s = rθ = θ
1 in the x y-plane consisting of all points ( x, y) which
2
2
satisfy the equation x + y = 1
...
1
...
So as the point ( x, y) goes around the
Figure 5
...
1
circle, its y-coordinate is sin θ
...
1
...
6 3 2
f (θ )
1
1
π
2
π
3
f (θ ) = sin θ
π
6
θ
2
2
x +y =1
Figure 5
...
2
0
1
0
π
6
π
3
π
2
2π
3
5π
6
π
θ
Graph of sine function based on y-coordinate of points on unit circle
We can extend the above picture to include angles from 0 to 2π radians, as in Figure 5
...
3
...
103
104
Chapter 5 • Graphing and Inverse Functions
§5
...
1
...
1
...
Recall from Section 1
...
So cos 0◦ has the same value as sin 90◦ , cos 90◦ has the
same value as sin 180◦ , cos 180◦ has the same value as sin 270◦ , and so on
...
1
...
1
...
1
105
y
8
6
4
2
−2π − 7π − 3π − 5π
4
2
4
−π − 3π
4
−π
2
−π
4
y = tan x
0
π
4
π
2
3π
4
π
x
5π
4
3π
2
7π
4
2π
−2
−4
−6
−8
Figure 5
...
6
Graph of y = tan x
Recall that the tangent is positive for angles in QI and QIII, and is negative in QII and
QIV, and that is indeed what the graph in Figure 5
...
6 shows
...
e
...
We can figure out
2
2
what happens near those angles by looking at the sine and cosine functions
...
And the closer
x
π
x gets to 2 , the larger tan x gets
...
2
Likewise, for x in QII very close to π , sin x is very close to 1 and cos x is negative and very
2
sin
close to 0, so the quotient tan x = cos x is a negative number that is very large, and it gets
x
larger in the negative direction the closer x gets to π
...
Similarly, we
2
π
π
get vertical asymptotes at x = − π , x = 32 , and x = − 32 , as in Figure 5
...
6
...
e
...
The graphs of the remaining trigonometric functions can be determined by looking at the
1
graphs of their reciprocal functions
...
We will get vertical asymptotes when sin x = 0,
namely at multiples of π: x = 0, ± π, ± 2π, etc
...
1
...
106
Chapter 5 • Graphing and Inverse Functions
§5
...
1
...
1
...
Note the vertical asymptotes at x = ± π , ± 32
...
y
4
y = sec x
3
2
1
−2π − 7π − 3π − 5π
4
2
4
−π − 3π
4
−π
2
−π
4
−1
0
π
4
π
2
3π
4
−2
−3
−4
Figure 5
...
8 Graph of y = sec x
π
x
5π
4
3π
2
7π
4
2π
Graphing the Trigonometric Functions • Section 5
...
Alternatively, we can
use the relation cot x = − tan ( x + 90◦ ) from Section 1
...
1
...
1
...
1
Draw the graph of y = − sin x for 0 ≤ x ≤ 2π
...
So reflecting the
graph of y = sin x around the x-axis gives us the graph of y = − sin x:
y
y = − sin x
1
x
0
π
4
π
2
3π
4
π
5π
4
3π
2
7π
4
2π
−1
Note that this graph is the same as the graphs of y = sin ( x ± π) and y = cos ( x + π )
...
1
It is worthwhile to remember the general shapes of the graphs of the six trigonometric
functions, especially for sine, cosine, and tangent
...
Many phenomena in nature exhibit sinusoidal
behavior, so recognizing the general shape is important
...
2
Draw the graph of y = 1 + cos x for 0 ≤ x ≤ 2π
...
So adding 1 to cos x moves the graph
of y = cos x upward by 1, giving us the graph of y = 1 + cos x:
y
2
y = 1 + cos x
1
0
x
π
4
π
2
3π
4
π
5π
4
3π
2
7π
4
2π
Exercises
For Exercises 1-12, draw the graph of the given function for 0 ≤ x ≤ 2π
...
y = − cos x
2
...
y = 2 − cos x
4
...
y = − tan x
6
...
y = 1 + sec x
8
...
y = 2 sin x
10
...
y = −2 tan x
12
...
We can extend the unit circle definition of the sine and cosine
functions to all six trigonometric functions
...
1
...
Identify each of the six trigonometric functions of θ with exactly
one of the line segments in Figure 5
...
10, keeping in mind that the
radius of the circle is 1
...
14
...
1
...
Q
y
R
S
P
θ
O
M
N
1
x
Figure 5
...
10
15
...
What is the geometric
interpretation of R α ( x, y)? Also, show that R −α (R α ( x, y)) = ( x, y) and R β (R α ( x, y)) = R α+β ( x, y)
...
2
109
5
...
1 how the graphs of the trigonometric functions repeat every 2π radians
...
First, recall that the domain of a function f ( x) is the set of all numbers x for which the
function is defined
...
The range of a function f ( x) is the set of all values that f ( x) can take over its domain
...
e
...
A function f ( x) is periodic if there exists a number p > 0 such that x + p is in the domain
of f ( x) whenever x is, and if the following relation holds:
f ( x + p ) = f ( x)
for all x
(5
...
If there is a smallest
such number p, then we call that number the period of the function f ( x)
...
3
The functions sin x, cos x, csc x, and sec x all have the same period: 2π radians
...
1 that the graphs of y = tan x and y = cot x repeat every 2π radians but they also repeat every π
radians
...
Example 5
...
2
...
Note that sin 2 x “goes twice as fast” as sin x
...
2
...
While sin x takes a full 2π radians to go through an entire cycle (the
4
largest part of the graph that does not repeat), sin 2 x goes through an entire cycle in just π radians
...
110
Chapter 5 • Graphing and Inverse Functions
§5
...
Since sin x has period 2π,1 we know that sin ( x + 2π) = sin x for all x
...
Now define
f ( x) = sin 2 x
...
We have to show
that p > 0 can not be smaller than π
...
That
is, assume that 0 < p < π, then show that this leads to some contradiction, and hence can not
be true
...
Then 0 < 2 p < 2π, and hence
sin 2 x = f ( x)
= f ( x + p)
(since p is the period of f ( x))
= sin 2( x + p)
= sin (2 x + 2 p)
for all x
...
e u = 2( u/2)), this means
that sin u = sin ( u + 2 p) for all real numbers u, and hence the period of sin x is as most 2 p
...
Why? Because the period of sin x is 2π > 2 p
...
The above may seem like a lot of work to prove something that was visually obvious from
the graph (and intuitively obvious by the “twice as fast” idea)
...
And the argument works for the other trigonometric functions as well
...
g
...
1 We will usually leave out the “radians” part when discussing periods from now on
...
2
111
Example 5
...
2
...
The graphs of both functions are
1
2x
y = cos 3 x
1
x
0
π
6
π
3
π
2
2π
3
5π
6
π
7π
6
4π
3
3π
2
5π 11π
3
6
2π
13π 7π
6
3
5π
2
8π 17π
3
6
3π
−1
Figure 5
...
2 Graph of y = cos 3 x and y = cos
19π 10π 7π 11π 23π
6
3
2
3
6
4π
1
2x
We know that −1 ≤ sin x ≤ 1 and −1 ≤ cos x ≤ 1 for all x
...
In this case, we call | A | the amplitude of the functions y = A sin x and y = A cos x
...
2
...
y
| A|
| A|
2| A|
x
0
π
4
π
2
3π
4
π
5π
4
3π
2
7π
4
=
| A |−(−| A |)
2
2π
| A|
−| A |
Figure 5
...
3
Amplitude =
max−min
2
= | A|
Not all periodic curves have an amplitude
...
Likewise, cot x, csc x, and sec x do not have
an amplitude
...
112
Chapter 5 • Graphing and Inverse Functions
§5
...
6
Find the amplitude and period of y = 3 cos 2 x
...
The graph is shown in Figure 5
...
4:
y
3
2
3
1
6
0
π
6
−1
π
3
π
2
2π
3
π
5π
6
x
7π
6
4π
3
3π
2
5π
3
11π
6
2π
3
−2
−3
y = 3 cos 2 x
Figure 5
...
4
Example 5
...
π
Solution: The amplitude of −3 sin 23 x is |−3 | = 3
...
It just shifts the entire graph upward by 2
...
2
2
2
= 3
...
2
...
2
...
2
113
Example 5
...
Solution: This is not a periodic function, since the angle that we are taking the sine of, x2 , is not a
linear function of x, i
...
is not of the form ax + b for some constants a and b
...
Can we say that
sin ( x2 ) is some constant times as fast as sin x ? No
...
This can be seen in
the graph of y = 2 sin ( x2 ), shown in Figure 5
...
6:2
2
1
...
5
0
-0
...
5
-2
0
π
2
π
3π
2
2π
x
Figure 5
...
6
y = 2 sin ( x2 )
Notice how the curve “speeds up” as x gets larger, making the “waves” narrower and narrower
...
Despite this, it appears that the function does have an amplitude,
namely 2
...
In the exercises you will be asked to find values of x such that 2 sin ( x2 ) reaches the maximum value
2 and the minimum value −2
...
Note: This curve is still sinusoidal despite not being periodic, since the general shape is still that of
a “sine wave”, albeit one with variable cycles
...
This will not always be the case
...
info
...
114
Chapter 5 • Graphing and Inverse Functions
§5
...
9
Find the amplitude and period of y = 3 sin x + 4 cos x
...
The period is still simple to determine: since sin x and cos x each repeat every 2π radians,
then so does the combination 3 sin x + 4 cos x
...
We can see
this in the graph, shown in Figure 5
...
7:
5
4
3
2
y
1
0
-1
-2
-3
-4
-5
0
π
2
π
3π
2
2π
5π
2
3π
7π
2
4π
x
Figure 5
...
7
y = 3 sin x + 4 cos x
The graph suggests that the amplitude is 5, which may not be immediately obvious just by looking
at how the function is defined
...
However, 3 sin x can never equal 3 for the same
x that makes 4 cos x equal to 4 (why?)
...
Let θ be the angle shown in the right
3
4
triangle in Figure 5
...
8
...
We can use this as follows:
y = 3 sin x + 4 cos x
= 5
3
5
sin x +
4
5
cos x
= 5 (cos θ sin x + sin θ cos x)
= 5 sin ( x + θ )
5
4
θ
3
Figure 5
...
8
(by the sine addition formula)
Thus, | y | = | 5 sin ( x + θ ) | = | 5 | · | sin ( x + θ ) | ≤ (5)(1) = 5, so the amplitude of y = 3 sin x + 4 cos x is 5
...
2
115
In general, a combination of sines and cosines will have a period equal to the lowest common multiple of the periods of the sines and cosines being added
...
9, sin x
and cos x each have period 2π, so the lowest common multiple (which is always an integer
multiple) is 1 · 2π = 2π
...
10
Find the period of y = cos 6 x + sin 4 x
...
The lowest common
π
2
= π
Thus, the period of y = cos 6 x + sin 4 x is π
...
2
...
5
1
y
0
...
5
-1
-1
...
2
...
9, since
we are not taking the cosine and sine of the same angle; we are taking the cosine of 6 x but the sine of
4 x
...
In Chapter 6, we will describe how to use a numerical computation program to show that the
maximum and minimum are ± 1
...
2204 × 10−16 )
...
90596111871578
...
2
Generalizing Example 5
...
So
a +b
a +b
y = a sin ω x + b cos ω x will have amplitude a2 + b2
...
We have seen how adding a constant to a function shifts the entire graph vertically
...
y
Consider a function of the form y = A sin ω x, where A
and ω are nonzero constants
...
Then the amplitude is A and the period is 2π
...
2
...
x
0
Now consider the function y = A sin (ω x − φ), where φ is
π
2π
ω
ω
some constant
...
Also, we know − A
ω
that the sine function goes through an entire cycle when its
Figure 5
...
10 y = A sin ω x
angle goes from 0 to 2π
...
So as ω x − φ goes from 0 to 2π, an entire cycle of the function y = A sin (ω x − φ)
will be traced out
...
ω
ω
and ends when
Thus, the graph of y = A sin (ω x − φ) is just the graph of y = A sin ω x shifted horizontally by
φ
, as in Figure 5
...
11
...
The amount
φ
ω
of the shift is called the phase shift of the graph
...
2
...
2
117
The phase shift is defined similarly for the other trigonometric functions
...
11
Find the amplitude, period, and phase shift of y = 3 cos (2 x − π)
...
2
...
The graph is shown in
period = π
3
2
amplitude = 3
1
0
−1
x
π
π
2
2π
3π
2
−2
−3
phase shift =
π
2
Figure 5
...
12
y = 3 cos (2 x − π)
Notice that the graph is the same as the graph of y = 3 cos 2 x shifted to the right by π , the amount
2
of the phase shift
...
12
Find the amplitude, period, and phase shift of y = −2 sin 3 x + π
...
Notice the negative
6
sign in the phase shift, since 3 x + π = 3 x − (−π) is in the form ω x − φ
...
2
...
2
...
2
In engineering two periodic functions with the same period are said to be out of phase if
their phase shifts differ
...
Periodic functions with the same period and the same phase shift are in phase
...
Then graph one
cycle of the function, either by hand or by using Gnuplot (see Appendix B)
...
y = 3 cos π x
2
...
y = − sin (5 x + 3)
4
...
y = 2 + cos (5 x + π)
6
...
y = 1 − cos (3π − 2 x)
8
...
y = 1 − tan (3π − 2 x)
10
...
y = 2 csc (2 x − 1)
12
...
For the function y = 2 sin ( x2 ) in Example 5
...
For the function y = 3 sin x + 4 cos x in Example 5
...
15
...
What are
the amplitude and period of this function?
16
...
Sketch one
cycle of both i ( t) and v( t) together on the same graph (i
...
on the same set of axes)
...
Repeat Exercise 16 with i ( t) the same as before but with v( t) = Vm sin ω t + π
...
Repeat Exercise 16 with i ( t) = − I m cos ω t − π and v( t) = Vm sin ω t − 56
...
Then graph one cycle of the
function, either by hand or by using Gnuplot
...
2
19
...
y = −5 sin 3 x + 12 cos 3 x
119
21
...
Find the amplitude of the function y = 2 sin ( x2 ) + cos ( x2 )
...
Graph one cycle using Gnuplot
...
y = sin 3 x − cos 5 x
24
...
y = 2 sin π x + 3 cos
π
3x
26
...
5 sin x sin 12 x
...
2
...
5 ∗ sin( x) ∗ sin(12 ∗ x)
0
...
5 ∗ sin( x)
y
0
...
5
-1
0
2π
π
3π
4π
x
Figure 5
...
14
Modulated wave y = 0
...
5 sin x
...
5
...
5
...
5 sin x form an amplitude envelope for the
wave (i
...
they enclose the wave)
...
4 to write this function as a sum
of sinusoidal curves
...
Use Gnuplot to graph the function y = x2 sin 10 x from x = −2π to x = 2π
...
)
1
28
...
2 to x = π
...
)
29
...
30
...
What happens at x = 0?
120
Chapter 5 • Graphing and Inverse Functions
§5
...
3 Inverse Trigonometric Functions
We have briefly mentioned the inverse ✄
trigonometric functions before, for example in Section
✄
✄
1
...
We will now define those inverse
functions and determine their graphs
...
We can write that rule as y = f ( x), where f is
y = f ( x)
the function (see Figure 5
...
1)
...
3
...
3
...
y
y
y = f ( x)
y = f ( x)
x
x
(a) f is a function
(b) f is not a function
Vertical rule for functions
Figure 5
...
2
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
...
Equivalently, f is
one-to-one if f ( x1 ) = f ( x2 ) implies x1 = x2
...
3
...
y
y
y = f ( x)
y = f ( x)
x
(a) f is one-to-one
x
(b) f is not one-to-one
Figure 5
...
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 , such that y = f ( x) if and only if f −1 ( y) = x
...
−1
Inverse Trigonometric Functions • Section 5
...
In other words,
f −1 ( f ( x)) = x
for all x in the domain of f , and
f (f
for all y in the range of f
...
However, we can restrict those functions to subsets of their domains
where they are one-to-one
...
3
...
In other words:
2 2
sin−1 (sin y) = y
sin (sin−1 x) = x
for − π ≤ y ≤
2
π
2
for −1 ≤ x ≤ 1
(5
...
3)
Example 5
...
Solution: Since − π ≤
2
π
4
≤ π , we know that sin−1 sin
2
π
4
=
π
4
, by formula (5
...
Example 5
...
π
π
Solution: Since 54 > π , we can not use formula (5
...
But we know that sin 54 = − 1
...
That
2
2
2
2
angle is y = − π , since
4
sin − π = − sin π = − 1
...
4
122
Chapter 5 • Graphing and Inverse Functions
§5
...
14 illustrates an important point: sin−1 x should always be a number between
− π and π
...
2
2
✄
sin
This why in Example 1
...
5 we got sin−1 (−0
...
Instead of an angle between 0◦ and 360◦ (i
...
0 to 2π radians) we got
an angle between −90◦ and 90◦ (i
...
− π to π radians)
...
The graph of y = sin−1 x is shown in Figure 5
...
5
...
y
y = sin−1 x
π
2
y = sin x
1
0
− π −1
2
x
1
π
2
−1
−π
2
y=x
Figure 5
...
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
...
3
...
In other words:
Inverse Trigonometric Functions • Section 5
...
4)
cos (cos
for −1 ≤ x ≤ 1
(5
...
3
...
Notice the symmetry about the
line y = x with the graph of y = cos x
...
3
...
15
Find cos−1 cos
π
3
...
4)
...
16
Find cos−1 cos
4π
3
...
e
...
π
1
we can not use formula (5
...
But we know that cos 43 = − 2
...
That
Thus, cos−1 cos
4π
3
=
2π
3
...
14 and 5
...
But that rule only applies
when the function f is one-to-one over its entire domain
...
That general rule, therefore, only holds for x in those small subsets in the
case of the inverse sine and inverse cosine
...
3
The inverse tangent function y = tan−1 x (sometimes called the arc tangent and denoted by y = arctan x) can be determined similarly
...
3
...
3
...
3
...
Notice that the vertical asymptotes for y = tan x become horizontal asymptotes for y = tan−1 x
...
y
3
y = tan x
2
π
2
1
−π
2
−π
4
y = tan−1 x
0
π
4
π
2
−1
y=x
−π
2
−2
−3
Figure 5
...
9
Graph of y = tan−1 x
x
Inverse Trigonometric Functions • Section 5
...
In other words:
2 2
tan−1 (tan y) = y
for − π < y <
2
tan (tan
for all real x
−1
x) = x
π
2
(5
...
7)
Example 5
...
Solution: Since − π ≤
2
π
4
≤ π , we know that tan−1 tan
2
π
4
π
=
4
, by formula (5
...
Example 5
...
Solution: Since π > π , we can not use formula (5
...
But we know that tan π = 0
...
That angle is y = 0
...
Example 5
...
1
1
Solution: Let θ = sin−1 − 4
...
2
2
Hence cos θ > 0
...
4
Note that we took the positive square root above since cos θ > 0
...
4
=
Example 5
...
1 − x2
Solution: When x = 0, the formula holds trivially, since
tan (sin−1 0) = tan 0 = 0 =
0
1 − 02
...
Let θ = sin−1 x
...
Draw a right triangle with an angle θ such that the opposite leg has length x and
the hypotenuse has length 1, as in Figure 5
...
10 (note that this is possible since
x
0 < x < 1)
...
By the Pythagorean Theorem, the adjacent leg has
length
1 − x2
...
1− x2
1
x
θ
1 − x2
Figure 5
...
10
If −1 < x < 0 then θ = sin−1 x is in QIV
...
(negative) in QIV
...
3
The inverse functions for cotangent, cosecant, and secant can be determined by looking at
their graphs
...
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)
...
8)
for all real x
(5
...
3
...
y
π
π
2
y = cot−1 x
π
− 34
−π
2
−π
4
0
Figure 5
...
11
π
4
π
2
x
3π
4
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 ≤ π , y = 0
...
2
csc−1 (csc y) = y
csc (csc−1 x) = x
π
≤ y≤ , y=0
2
2
for | x | ≥ 1
for −
π
sec−1 (sec y) = y
for 0 ≤ y ≤ π, y =
sec (sec−1 x) = x
for | x | ≥ 1
π
2
(5
...
11)
(5
...
13)
It is also common to call cot−1 x, csc−1 x, and sec−1 x the arc cotangent, arc cosecant,
and arc secant, respectively, of x
...
3
...
3
y
127
y
π
π
2
y = csc−1 x
π
2
x
−1 0
y = sec−1 x
1
−π
2
x
−1 0
(a) Graph of y = csc−1 x
1
(b) Graph of y = sec−1 x
Figure 5
...
12
Example 5
...
Solution: Let θ = cot−1 x
...
5, we have
tan
π
2
− θ = − tan θ − π = cot θ = cot (cot−1 x) = x ,
2
by formula (5
...
So since tan (tan−1 x) = x for all x, this means that tan (tan−1 x) = tan π − θ
...
Now, we know that 0 < cot−1 x < π, so − π < π − cot−1 x <
2
2
2
π
π
−1
x is in the restricted subset on which the tangent function is one-to-one
...
e
...
2
2
Example 5
...
Then
tan (tan−1 a + tan−1 b) =
tan A + tan B
, let A = tan−1 a and B =
1 − tan A tan B
tan (tan−1 a) + tan (tan−1 b)
1 − tan (tan−1 a) tan (tan−1 b)
a+b
by formula (5
...
However, recall that − π < tan−1 x <
2
in particular, we must have
−π
2
< tan−1 1a+b
−ab
the interval − π , π
...
true when − π < tan−1 a
2
for all real numbers x
...
892547 >
And we see that tan−1
π
2
π
2
≈ 1
...
1+2
−1
−1
1−(1)(2) = tan (−3) = −1
...
2
+ tan−1 2
...
3
Exercises
For Exercises 1-25, find the exact value of the given expression in radians
...
tan−1 1
2
...
tan−1 0
4
...
cos−1 (−1)
6
...
sin−1 1
8
...
sin−1 0
10
...
sin−1 sin
4π
3
π
12
...
cos−1 cos
6π
5
16
...
csc−1 csc − π
9
23
...
sec−1 sec
4
5
13
...
cos−1 cos − 10
π
17
...
sin−1
π
7
21
...
cot−1 cot
5
13
4π
3
22
...
tan−1
3
5
+ cot−1
3
5
For Exercises 26-33, prove the given identity
...
cos (sin−1 x) =
27
...
sin−1 x + cos−1 x =
π
2
29
...
sin−1 (− x) = − sin−1 x
32
...
cos−1 (− x) + cos−1 x = π
33
...
In Example 5
...
Does the formula tan (tan−1 a + tan−1 b) =
hold? Explain your answer
...
Show that tan−1
1
3
+ tan−1
1
5
= tan−1
4
7
1
4
+ tan−1
2
9
= tan−1
1
2
for x > 0
a+b
1 − ab
does not always
a+b
, which was part of that example, always
1 − ab
...
Show that tan−1
π
2
...
Figure 5
...
13 shows three equal squares lined up against each other
...
(Hint: Consider the tangents of the angles
...
3
...
Sketch the graph of y = sin−1 2 x
...
Write a computer program to solve a triangle in the case where you are given three sides
...
Note that since most computer languages use radians for their
inverse trigonometric functions, you will likely have to do the conversion from radians to degrees
yourself in the program
...
1 Solving Trigonometric Equations
An equation involving trigonometric functions is called a trigonometric equation
...
75 ,
which we encountered in Chapter 1, is a trigonometric equation
...
In this section we
will be concerned with finding the most general solution to such equations
...
75
...
87◦
...
e
...
Thus, there are many other possible
answers for the value of A , namely 36
...
87◦ − 180◦ , 36
...
87◦ − 360◦ ,
36
...
We can write this in a more compact form:
A = 36
...
This is the most general solution to the equation
...
” is omitted since it is usually understood that k varies over all integers
...
6435 + π k
for k = 0, ± 1, ± 2,
...
1
Solve the equation 2 sin θ + 1 = 0
...
Using the ✂ −1 ✁calculator button in degree mode gives
2
us θ = −30◦ , which is in QIV
...
That is, sin 210◦ = − 2
...
In radians, the solution is:
θ = −
π
6
+ 2π k
and
7π
+ 2π k
6
for k = 0, ± 1, ± 2,
...
129
130
Chapter 6 • Additional Topics
§6
...
2
Solve the equation 2 cos2 θ − 1 = 0
...
But notice that the above angles differ by multiples of π
...
k
Example 6
...
Solution: Isolating sec θ gives us
sec θ =
1
2
cos θ =
⇒
1
= 2,
sec θ
which is impossible
...
Example 6
...
Solution: The idea here is to use identities to put everything in terms of a single trigonometric
function:
cos θ = tan θ
sin θ
cos θ
cos2 θ = sin θ
cos θ =
1 − sin2 θ = sin θ
0 = sin2 θ + sin θ − 1
The last equation looks more complicated than the original equation, but notice that it is actually a
quadratic equation: making the substitution x = sin θ , we have
x2 + x − 1 = 0
⇒
x =
−1 ±
−1 ±
1 − (4) (−1)
=
2 (1)
2
5
= −1
...
618
by the quadratic formula from elementary algebra
...
618 < −1, so it is impossible that sin θ =
x = −1
...
Thus, we must have sin θ = x = 0
...
Hence, there are two possible solutions: θ = 0
...
475 rad around the y-axis in QII
...
666 + 2π k
and
2
...
Solving Trigonometric Equations • Section 6
...
5
Solve the equation sin θ = tan θ
...
So since the above angles are multiples of π, and every multiple of 2π is a
multiple of π, we can combine the two answers into one for the general solution:
for k = 0, ± 1, ± 2,
...
6
Solve the equation cos 3θ =
1
2
...
So since cos−1 1 = π , there are two possible solutions for 3θ : 3θ = π in QI and its
2
3
3
reflection −3θ = − π around the x-axis in QIV
...
+ 2π k
So dividing everything by 3 we get the general solution for θ :
θ = ±
π
9
+
2π
k
3
for k = 0, ± 1, ± 2,
...
7
Solve the equation sin 2θ = sin θ
...
132
Chapter 6 • Additional Topics
§6
...
8
Solve the equation 2 sin θ − 3 cos θ = 1
...
Take the coefficients 2 and
3 of sin θ and − cos θ , respectively, in the above equation and make them the legs
of a right triangle, as in Figure 6
...
1
...
The leg with length 3 > 0 means that the angle φ is above the x-axis, and the leg with
length 2 > 0 means that φ is to the right of the y-axis
...
The
hypotenuse has length 13 by the Pythagorean Theorem, and hence cos φ = 2 and
sin θ =
3
...
1
...
281
θ = φ + 0
...
281 = 2
...
861
Now, since cos φ = 2 and φ is in QI, the most general solution for φ is φ = 0
...
So since we needed to add multiples of 2π to the solutions 0
...
861 anyway, the most
general solution for θ is:
θ = 0
...
281 + 2π k
and
0
...
861 + 2π k
θ = 1
...
844 + 2π k
⇒
for k = 0, ± 1, ± 2,
...
8 if the equation had been 2 sin θ + 3 cos θ = 1 then we still would
have used a right triangle with legs of lengths 2 and 3, but we would have used the sine
addition formula instead of the subtraction formula
...
1
...
2 cos θ + 1 = 0
2
3
...
2 cos θ − sin θ = 1
5
...
2 cos2 θ + 3 sin θ = 0
7
...
tan θ + cot θ = 2
9
...
2 sin θ − 3 cos θ = 0
11
...
3 sin θ − 4 cos θ = 1
Numerical Methods in Trigonometry • Section 6
...
2 Numerical Methods in Trigonometry
We were able to solve the trigonometric equations in the previous section fairly easily, which
in general is not the case
...
(6
...
2
...
The graphs of y = cos x and y = x
intersect somewhere between x = 0 and x = 1, which means that there is an x in the interval
[0, 1] such that cos x = x
...
2
...
Instead, we have to resort to numerical methods, which provide ways of getting successively
better approximations to the actual solution(s) to within any desired degree of accuracy
...
Many
of the methods require calculus, but luckily there is a method which we can use that requires
just basic algebra
...
e
...
A derivation of the secant method is beyond the scope of
this book,1 but we can state the algorithm it uses to solve f ( x) = 0:
1 For an explanation of why the secant method works, see pp
...
R ALSTON AND P
...
, New York: McGraw-Hill Book Co
...
134
Chapter 6 • Additional Topics
§6
...
Pick initial points x0 and x1 such that x0 < x1 and f ( x0 ) f ( x1 ) < 0 (i
...
the solution is
somewhere between x0 and x1 )
...
For n ≥ 2, define the number xn by
x n = x n −1 −
( x n −1 − x n −2 ) f ( x n −1 )
f ( x n −1 ) − f ( x n −2 )
(6
...
3
...
will approach the solution x as we go through more iterations, getting as close as desired
...
The solution
to that equation is the root of the function f ( x) = cos x − x
...
So pick x0 = 0 and x1 = 1
...
4597,
so that f ( x0 ) f ( x1 ) < 0 (we are using radians, of course)
...
4597)
= 1 −
−0
...
6851 ,
x2 = x1 −
( x2 − x1 ) f ( x2 )
f ( x2 ) − f ( x1 )
(0
...
6851)
= 0
...
6851) − f (1)
(0
...
0893)
= 0
...
0893 − (−0
...
7363 ,
x3 = x2 −
and so on
...
A better way to implement the algorithm is with a computer
...
1 below shows the code
(secant
...
2
135
Listing 6
...
java
1
2
import java
...
*;
public class secant {
public static void main (String[] args) {
3
4
double x0 =
Double
...
parseDouble(args[1]);
6
double x = 0;
7
double error = 1
...
compare(Math
...
out
...
out
...
cos(x) - x;
24
}
25
26
}
Lines 4-5 read in x0 and x1 as input parameters to the program
...
Line 7 sets the maximum error ǫ error to be 1
...
That is, our final answer will be
within that (tiny!) amount of the real solution
...
e
...
, x10 to the real solution, though in Line 9 we check to see if the two
previous approximations differ by less than the maximum error
...
Line 10 is the main step in the algorithm, creating xn from xn−1 and xn−2
...
Lines 18-20 set the number of decimal places to show in the final answer to 50 (the default
is 16) and then print the answer
...
136
Chapter 6 • Additional Topics
§6
...
java
java secant 0 1
x2 = 0
...
736298997613654
x4 = 0
...
7390851121274639
x6 = 0
...
7390851332151607
x8 = 0
...
73908513321516067229310920083662495017051696777344
Notice that the program only got up to x8 , not x10
...
0 × 10−50 ) to stop at x8 and call
that our solution
...
Does that number look familiar? It should, since it is the answer to Exercise 11 in Section
4
...
That is, when taking repeated cosines starting with any number (in radians), you eventually start getting the above number repeatedly after enough iterations
...
Figure 6
...
2 gives an idea of why
...
8
y
0
...
4
y=x
0
...
2
...
73908513321516
...
This number x is called an attractive fixed
Numerical Methods in Trigonometry • Section 6
...
No matter where you start, you end up getting “drawn” to it
...
2
...
e
...
Recall in Example 5
...
2 that we claimed that the maximum and minimum of
the function y = cos 6 x + sin 4 x were ± 1
...
We can show this by
using the open-source program Octave
...
Finding the maximum of f ( x) is the same as
finding the minimum of − f ( x) then multiplying by −1 (why?)
...
The command sqp(3,’f’) says to use x = 3 as a first approximation of the number x
where f ( x) is a minimum
...
90596111871578
ans = 2
...
65792064609274 and that the minimum is −1
...
To find the maximum of f ( x), we find the minimum of − f ( x)
and then take its negative
...
octave:4> function y = f(x)
> y = -cos(6*x) - sin(4*x)
> endfunction
octave:5> sqp(2,’f’)
y = -1
...
05446832062993
The output says that the maximum occurs when x = 2
...
90596111871578) = 1
...
Recall from Section 2
...
However, you can get around this problem by
using computer software capable of handling numbers with a high degree of precision
...
gnu
...
2
modern computer programming languages have this capability
...
4 Below we show how to get accuracy up to 50 decimal places using
Heron’s formula for the triangle in Example 2
...
4, by using the python
interactive command shell:
>>> from decimal import *
>>> getcontext()
...
9999979")
>>> c = Decimal("0
...
sqrt()
0
...
)
Notice in this case that we do get the correct answer; the high level of precision eliminates
the rounding errors shown by many calculators when using Heron’s formula
...
It can be run on your own computer, but it can also be run through a web
interface: go to http://sagenb
...
For example, to find the
solution to cos x = x in the interval [0, 1], enter these commands in the worksheet textfield:
x = var(’x’)
find_root(cos(x) == x, 0,1)
Click the evaluate link to display the answer: 0
...
One obvious solution to the equation 2 sin x = x is x = 0
...
0 × 10−20
...
1 (only one line needs to be
changed!)
...
2
...
3
...
3 Available for free at http://www
...
org
4 Other languages have similar capability, e
...
the BigDecimal class in Java
...
sagemath
...
Complex Numbers • Section 6
...
3 Complex Numbers
There is no real number x such that x2 = −1
...
Thus, i 2 = −1, and
hence i = −1
...
The real number a is called the real part of the complex number
a + bi , and bi is called its imaginary part
...
e
...
If b = 0 then a + bi = a + 0 i = a (since 0 i is defined as 0), so that every real
number is a complex number
...
Before exploring that correspondence further, we will first state some fundamental properties of and operations on complex numbers:
Let a + bi and c + di be complex numbers
...
a + bi = c + di if and only if a = c and b = d (i
...
the real parts are equal and the
imaginary parts are equal)
2
...
e
...
(a + bi ) − ( c + di ) = (a − c) + ( b − d ) i
4
...
(a + bi ) (a − bi ) = a2 + b2
6
...
The last three items can be derived by treating the multiplication and
division of complex numbers as you would normally treat factors of real numbers:
(a + bi ) ( c + di ) = a ( c + di ) + bi ( c + di )
= ac + adi + bci + bdi 2 = ac + adi + bci + bd (−1)
= (ac − bd ) + (ad + bc) i
6 Especially in electrical engineering, physics, and various fields of mathematics
...
3
The fifth item is a special case of the multiplication formula:
(a + bi ) (a − bi ) = ((a)(a) − ( b)(− b)) + ((a)(− b) + ( b)(a)) i
= (a2 + b2 ) + (−ab + ba) i = (a2 + b2 ) + 0 i
= a2 + b 2
The sixth item comes from using the previous items:
a + bi c − di
a + bi
=
·
c + di
c + di c − di
(ac − b(− d )) + (a(− d ) + bc) i
=
c2 + d 2
(ac + bd ) + ( bc − ad ) i
=
c2 + d 2
The conjugate a + bi of a complex number a + bi is defined as a + bi = a − bi
...
So for a complex number z = a + bi ,
z z = a2 + b2 and thus we can define the modulus of z to be z z = a2 + b2 , which we denote
by | z |
...
9
Let z1 = −2 + 3 i and z2 = 3 + 4 i
...
Solution: Using our rules and definitions, we have:
z1 + z2 = (−2 + 3 i ) + (3 + 4 i )
= 1 + 7i
z1 − z2 = (−2 + 3 i ) − (3 + 4 i )
= −5 − i
z1 z2 = (−2 + 3 i ) (3 + 4 i )
= ((−2)(3) − (3)(4)) + ((−2)(4) + (3)(3)) i
= −18 + i
−2 + 3 i
z1
=
z2
3 + 4i
(−2)(3) + (3)(4) + ((3)(3) − (−2)(4)) i
=
32 + 42
17
6
+
i
=
25
25
| z1 | =
=
| z2 | =
= 5
(−2)2 + 32
13
32 + 42
Complex Numbers • Section 6
...
3
...
y
y
( x, y) = ( r cos θ , r sin θ )
z = x + yi = r cos θ + ( r sin θ ) i
r
θ
0
r
θ
x
0
(a) Point ( x, y)
x
(b) Complex number z = x + yi
Figure 6
...
1
Let z = x + yi be a complex number
...
3
...
The distance r from z to the origin is,
by the Pythagorean Theorem, r = x2 + y2 , which is just the modulus of z
...
3
...
We call this angle θ the argument of z
...
3)
x2 + y2 and
θ = the argument of z
...
4)
In the special case z = 0 = 0 + 0 i , the argument θ is undefined since r = | z | = 0
...
Note also that for z = x + yi with
r = | z |, θ must satisfy
y
y
tan θ = x , cos θ = x , sin θ = r
...
3
Example 6
...
Solution: Let z = −2 − i = x + yi , so that x = −2 and y = −1
...
3
...
So since tan θ = x = −1 = 2 , we have
−2
◦
θ = 206
...
Also,
r =
x 2 + y2 =
(−2)2 + (−1)2 =
θ
2
1
5
...
3
...
6◦ + i sin 206
...
6◦
...
Then
z1 z2 = r 1 r 2 (cos (θ1 + θ2 ) + i sin (θ1 + θ2 )) , and
r1
z1
=
(cos (θ1 − θ2 ) + i sin (θ1 − θ2 )) if z2 = 0
...
5)
(6
...
And
z1
r 1 (cos θ1 + i sin θ1 )
=
z2
r 2 (cos θ2 + i sin θ2 )
r 1 cos θ1 + i sin θ1 cos θ2 − i sin θ2
·
·
=
r 2 cos θ2 + i sin θ2 cos θ2 − i sin θ2
r 1 (cos θ1 cos θ2 + sin θ1 sin θ2 ) + i (sin θ1 cos θ2 − cos θ1 sin θ2 )
=
·
r2
cos2 θ2 + sin2 θ2
r1
=
(cos (θ1 − θ2 ) + i sin (θ1 − θ2 ))
r2
by the subtraction formulas for sine and cosine, and since cos2 θ2 + sin2 θ2 = 1
...
5) and (6
...
This makes working with complex
numbers in trigonometric form fairly simple
...
3
143
Example 6
...
Find z1 z2 and
z1
z2
...
5) and (6
...
z2
For the special case when z1 = z2 = z = r (cos θ + i sin θ ) in formula (6
...
e
...
1
...
(6
...
So by De Moivre’s Theorem and
formula (6
...
8
7 Named after the French statistician and mathematician Abraham de Moivre (1667-1754)
...
See pp
...
V
...
, New York:
McGraw-Hill Book Co
...
144
Chapter 6 • Additional Topics
§6
...
12
Find (1 + i )10
...
We can use De Moivre’s Theorem to find the n th roots of a complex number
...
Let z = r (cos θ + i sin θ )
...
Now let w = r 0 (cos θ0 + i sin θ0 )
be an n th root of z
...
will repeat for k ≥ n, we get the following formula for the
For any nonzero complex number z = r (cos θ + i sin θ ) and positive integer n, the n
distinct n th roots of z are
r 1/n cos
θ + 360◦ k
n
+ i sin
θ + 360◦ k
(6
...
, n − 1
...
The number r 1/n in the above formula
is the usual real n th root of the real number r = | z |
...
13
Find the three cube roots of i
...
3
145
y
Notice from Example 6
...
3
...
We see that consec3
3
i
i
120◦
2 +2
utive cube roots are 120◦ apart
...
n
−i
In higher mathematics the Fundamental Theorem of Algebra states that every polynomial of degree n with complex
Figure 6
...
3
coefficients has n complex roots (some of which may repeat)
...
For example,
the square roots of 1 are ± 1, and the square roots of −1 are ± i
...
1
...
(2 + 3 i ) − (−3 − 2 i )
3
...
(2 + 3 i )/(−3 − 2 i )
5
...
(2 + 3 i ) − (−3 − 2 i )
7
...
|−3 + 2 i |
9
...
i 5
13
...
i 9
10
...
i 6
14
...
i 2009
For Exercises 17-24, prove the given identity for all complex numbers
...
( z) = z
21
...
z1 + z2 = z1 + z2
19
...
z1 z2 = z1 z2
22
...
| z1 z2 | = | z1 | | z2 |
24
...
25
...
−3 − 2 i
27
...
− i
29
...
−1
31
...
For Exercises 32-35, calculate the given number
...
3 (cos 14◦ + i sin 14◦ ) · 2 (cos 121◦ + i sin 121◦ )
33
...
[3 (cos 14◦ + i sin 14◦ )]−4
35
...
Find the three cube roots of − i
...
Find the three cube roots of 1 + i
...
Find the three cube roots of 1
...
Find the three cube roots of −1
...
Find the five fifth roots of 1
...
Find the five fifth roots of −1
...
Find the two square roots of −2 + 2 3 i
...
Prove that if z is an n th root of a real number a, then so is z
...
)
146
Chapter 6 • Additional Topics
§6
...
4 Polar Coordinates
y
Suppose that from the point (1, 0) in the
x y-coordinate plane we draw a spiral around
the origin, such that the distance between
any two points separated by 360◦ along the
←
1→
spiral is always 1, as in Figure 6
...
1
...
x
0
1
2
3
However, this spiral would be simple to
describe using the polar coordinate system
...
4
...
We call
the pair ( r, θ ) the polar coordinates of P ,
Figure 6
...
1
and the positive x-axis is called the polar
axis of this coordinate system
...
, so (unlike
for Cartesian coordinates) the polar coordinates of a point are not unique
...
4
...
4
...
That is, the ray OP is drawn in the opposite direction from the
angle θ , as in Figure 6
...
3
...
You may be familiar with graphing paper, for plotting points or functions given in Cartesian coordinates (sometimes also called rectangular coordinates)
...
Similar graphing paper exists for plotting points and functions in polar
coordinates, similar to Figure 6
...
4
...
4
105◦
90◦
147
75◦
120◦
60◦
135◦
45◦
30◦
150◦
165◦
15◦
180◦
0◦
O
345◦
195◦
330◦
210◦
315◦
225◦
300◦
240◦
255◦
Figure 6
...
4
270◦
285◦
Polar coordinate graph
The angle θ can be given in either degrees or radians, whichever is more convenient
...
The reason is
that, unlike degrees, radians can be considered “unitless” (as we mentioned in Chapter 4)
...
For example, if a function in polar coordinates is written as r = 2 θ , then r
would have the same units as θ
...
148
Chapter 6 • Additional Topics
§6
...
14
Express the spiral from Figure 6
...
1 in polar coordinates
...
The goal is to find some equation involving r and θ that describes
the spiral
...
...
⇒
r = 2
r = 1+k
for k = 0, 1, 2,
...
So for
any θ ≥ 0,
θ
θ
θ = 2π k ⇒ k =
⇒ r = 1+k = 1+
...
The graph is shown in Figure 6
...
5, along
2π
with the Gnuplot commands to create the graph
...
4
...
Polar Coordinates • Section 6
...
4
...
For a point with polar coordinates ( r, θ ) and
Cartesian coordinates ( x, y):
Polar to Cartesian:
r
y
θ
x = r cos θ
O
(6
...
4
...
10)
Note that in formula (6
...
Also, if x = 0 and y = 0 then
y
the two possible solutions for θ in the equation tan θ = x are in opposite quadrants (for
0 ≤ θ < 2π)
...
e
...
x2 + y2 (i
...
Example 6
...
9) with r = 2 and θ = 30◦ , we get:
( x, y) = ( r cos θ , r sin θ ) = (2 cos 30◦ , 2 sin 30◦ ) = 2 ·
3
2 ,2
·
1
2
( x, y) =
⇒
3, 1
(b) Using formula (6
...
9) with r = −1 and θ = 5π/3, we get:
( x, y) = ( r cos θ , r sin θ ) = −1 cos
5π
5π
3 , −1 sin 3
= −1 ·
1
2 , −1
·
− 3
2
⇒
( x, y) = − 1 ,
2
Example 6
...
10) with x = 3 and y = 4, we get:
tan θ =
4
y
=
x
3
⇒
θ = 53
...
13◦
Since θ = 53
...
Thus, ( r, θ ) = (5, 53
...
Note that if we had used θ = 233
...
13◦ )
...
4
(b) Using formula (6
...
Thus, ( r, θ ) = (5 2, 225◦ )
...
Example 6
...
Solution: This is just the equation of a circle of radius 3 centered at the origin
...
x 2 + y2 =
Example 6
...
Solution: This is the equation of a circle of radius 4 centered at the point (0, 4)
...
Thus, the equation is r = 8 sin θ
...
19
Write the equation y = x in polar coordinates
...
So when x = 0, we know that y = 0
...
Thus, the equation is θ = 45◦
...
4
151
Example 6
...
2
1
(6
...
So write
x1 = r 1 cos θ1
x2 = r 2 cos θ2
y1 = r 1 sin θ1
y2 = r 2 sin θ2
...
Thus, by
the Cartesian coordinate distance formula,
d 2 = ( x1 − x2 )2 + ( y1 − y2 )2
= ( r 1 cos θ1 − r 2 cos θ2 )2 + ( r 1 sin θ1 − r 2 sin θ2 )2
= r 2 cos2 θ1 − 2 r 1 r 2 cos θ1 cos θ2 + r 2 cos2 θ2 + r 2 sin2 θ1 − 2 r 1 r 2 sin θ1 sin θ2 + r 2 sin2 θ2
1
2
1
2
= r 2 (cos2 θ1 + sin2 θ1 ) + r 2 (cos2 θ2 + sin2 θ2 ) − 2 r 1 r 2 (cos θ1 cos θ2 + sin θ1 sin θ2 )
1
2
d 2 = r 2 + r 2 − 2 r 1 r 2 cos (θ1 − θ2 ) ,
1
2
so the result follows by taking square roots of both sides
...
17 we saw that the equation x2 + y2 = 9 in Cartesian coordinates could be
expressed as r = 3 in polar coordinates
...
In general, polar coordinates are useful
in situations when there is symmetry about the origin (though there are other situations),
which arise in many physical applications
...
1
...
(−4, 3π)
3
...
(6, 90◦ )
5
...
6
...
(−1, −3)
8
...
(4, −2)
10
...
11
...
y = − x
13
...
Graph the function r = 1 + 2 cos θ in polar coordinates
...
3 x2 + 4 y2 − 6 x = 9
Appendix A
Answers and Hints to Selected Exercises
Chapter 1
13
...
1 (p
...
sin A = 5/ 106, cos A = 9/ 106,
csc A = 106/5, sec A = 106/9, cot A = 9/5
1
...
A = 52◦ , B = 104◦ 5
...
sin A = 40/7, cos A = 3/7,
7
...
0
...
989◦
tan A = 40/3, csc A = 7/ 40, cot A = 3/ 40
9
...
111
...
cos 3◦ 21
...
csc 13◦
15
...
sin 77◦ 27
...
Hint: Draw a
sided figure inside the circle parallel?
right triangle with an acute angle A
...
Hint: Draw two right triangles whose hypotenuses are the same length
...
2 (p
...
(a) 13/4 (b) 4 3/ 13 (c) 3/ 13
1
...
3 (p
...
102
...
241
...
274 ft 7
...
sin A = 7/25, cos A = 24/25, tan A = 7/24,
mi 9
...
476 in 11
...
955 in 13
...
4866
csc A = 25/7, sec A = 25/24, cot A = 24/7;
in 14
...
c = 13, A = 22
...
4◦ 17
...
28, c = 2
...
b = 6
...
sin A = 9/41, cos A = 40/41, tan A = 9/40,
6
...
a = 6
...
84, A = 64◦ 23
...
(a) 0
...
112 cm 27
...
26◦
7
...
1379
...
2613 mi
csc A = 10, sec A = 10/3, cot A = 3;
sin B = 3/ 10, cos B = 1/ 10, tan B = 3,
Section 1
...
31)
csc B = 10/3, sec B = 10, cot B = 1/3
9
...
QII 3
...
negative y-axis
csc A = 6/5, sec A = 6/ 11, cot A = 11/5;
7
...
QIV 11
...
QI,
QIV 15
...
43◦ 19
...
sin B = 11/6, cos B = 5/6, tan B = 11/5,
csc B = 6/ 11, sec B = 6/5, cot B = 5/ 11
85◦
23
...
cos A = 7/4, tan A = 3/ 7, csc A = 4/3, sin θ = − 3/2 and tan θ = 3
sec A = 4/ 7, cot A = 7/3
25
...
4 (p
...
cos θ = 3/2 and tan θ = 1/ 3;
1
...
55 3
...
21 5
...
905
cos θ = − 3/2 and tan θ = −1/ 3
4
7
...
21 9
...
cos θ = ±1 and tan θ = 0
quadrilateral into four triangles; also, con31
...
23)-(2
...
33
...
sin θ = 5/13 and cos θ = 12/13;
Section 2
...
64)
sin θ = −5/13 and cos θ = −12/13 37
...
No
1
...
63, r = 0
...
R = 3
...
36
5
...
18, r = 1
...
(c) Twice as large
Section 1
...
37)
(d) Hint: Bisect each angle
...
(a) 328◦ (b) 148◦ (c) 212◦
(b) 68◦ (c) 292◦ 7
...
38◦ , 218◦ 13
...
D = a2 +b2 , a2 +b 2
b
3
...
65◦ , 295◦
Section 3
...
70)
Chapter 2
Section 2
...
43)
1
...
4, c = 15
...
a = 9
...
7, C = 95◦ 5
...
1, B = 136
...
5◦
7
...
b = 24
...
9 , C = 70
...
9, B = 20
...
9◦ 11
...
5
...
86 cm 16
...
Section 2
...
49)
Chapter 3
1
...
Hint: See Example
3
...
19
...
2 (p
...
sin ( A + B) = 1020 , cos ( A + B) = − 1189 ,
1189
tan ( A + B) = − 1020 4
...
2 − 3 15
...
1
...
6, B = 40
...
1 3
...
9◦ , b = 8
...
1◦ 5
...
Section 3
...
81)
4
...
91 cm 9
...
5◦ , 59◦ , 70
...
7 cm 15
...
Hint: Is sin A + cos A always positive? 11
...
Section 2
...
53)
1
...
1◦ , B = 40
...
6
47
...
2, C = 72
...
No
11
...
Section 3
...
86)
3
...
Hint: One way to do this is with the Law
6
...
Another way is with the Law of
Sines
...
1 (p
...
π/45
174◦
3
...
−3π/5
7
...
2 (p
...
9
...
11π in 5
...
94 in
7
...
86 ft 8
...
18 9
...
26
11
...
392 and 9
...
3
...
3 (p
...
max
...
min
...
15
...
5, period = π 17
...
phase 18
...
amplitude = 34,
period = 2 21
...
2π 25
...
amplitude envelope: y = ± x2 29
...
3 (p
...
π/4 3
...
π 7
...
0
11
...
π/7 15
...
π/6
19
...
12/13 23
...
π/2
1
...
512 cm2 3
...
5 m2 5
...
1 cm2
Chapter 6
7
...
π/2 cm2 11
...
017 cm2 13
...
46 15
...
17 17
...
522 m2
Section 6
...
132)
19
...
34 + π k 3
...
± π + π k
doubled
...
−0
...
963 + 2π k 9
...
2πk
Section 4
...
102)
3
1
...
5 rad/sec
3
...
6 m/sec, ω = 0
...
ν = 3
...
875 rad/sec
7
...
375 rad 9
...
33 rpm
11
...
84 in/sec
Section 6
...
138)
1
...
89549426703398093962
Section 6
...
145)
1
...
−13 i 5
...
i
9
...
i 13
...
i
17
...
Then z = a − bi , so
Section 5
...
108)
( z) = a − bi = a + bi = z
...
Hint: Use Exercise 20
...
13 cis 56
...
2 cis 315◦
13
...
cis 0◦ 33
...
1
...
2 cis 15◦ , 2 cis 135◦ , 2 cis 255◦
Section 5
...
118)
39
...
cis 36◦ , cis 108◦ ,
2
2
1
...
amplitude = 1, period = 2π/5, phase shift =
−3/5 5
...
4 (p
...
amplitude = 1, period = π,
1
...
( 3, −1)
phase shift = 3π/2
◦
9
...
(−1/ 2, −1/ 2) 7
...
6 )
◦
shift = 3π/2 11
...
(2 5, 333
...
r = 6 cos θ
13
...
r = 3/(2 − cos θ )
riod = π, phase shift = 1/2
Chapter 5
Appendix B
Graphing with Gnuplot
Gnuplot is a free, open-source software package for producing a variety of graphs
...
Below is a very brief tutorial on how to use
Gnuplot to graph trigonometric functions
...
Go to http://www
...
info/download
...
For Windows, you should download the setup file
with a name such as gp460-win32-setup
...
6
...
All the examples
discussed here will assume at least version 4
...
0, though they should work with earlier
4
...
2
...
For example, in Windows you would run the setup file you
downloaded in Step 1, which installs Gnuplot in the C:\Program Files\gnuplot folder by
default
...
3
...
info/documentation
...
RUNNING GNUPLOT
1
...
exe from the bin folder where you installed Gnuplot (the default location is C:\Program Files\gnuplot\bin\wgnuplot
...
In Linux, just type gnuplot in
a terminal window
...
You should now get a Gnuplot terminal with a gnuplot> command prompt
...
In Linux it
will appear in the terminal window where the gnuplot command was run
...
” option
...
ini)
...
At the gnuplot> command prompt you can now run graphing commands, which we will
now describe
...
To specify an x range, use an expression of the form [a : b], for
some numbers a < b
...
To specify an x range and a y range, use an expression of the form [a : b][ c : d ], for some
numbers a < b and c < d
...
Function definitions use the x variable in combination with mathematical operators, listed
below:
Appendix B:
Symbol
+
−
*
/
**
exp( x)
log( x)
sin( x)
cos( x)
tan( x)
Operation
Addition
Subtraction
Multiplication
Division
Power
ex
ln x
sin x
cos x
tan x
Graphing with Gnuplot
Example
2+3
3−2
2*3
4/2
2**3
exp(2)
log(2)
sin(pi/2)
cos(pi)
tan(pi/4)
157
Result
5
1
6
2
23 = 8
e2
ln 2
1
−1
1
Note that we use the special keyword “pi” to denote the value of π
...
1
...
8
0
...
4
0
...
2
-0
...
6
-0
...
To get the x-axis labels with fractions of π,
you need to modify the terminal setting
...
To display it, use this command before the
plot command:
set zeroaxis
Also, to label the axes, use these commands:
set xlabel "x"
set ylabel "y"
The default sample size for plots is 100 units, which can result in jagged edges if the curve
is complicated
...
5
1
y
0
...
5
-1
-1
...
If you are using the
default wxt terminal then select Print near the top of the main Gnuplot window and enter
png in the Terminal type? textfield, then hit OK to get the Print Setup dialog
...
”, and enter png in the Terminal type? textfield, hit
OK
...
” option and enter a filename (say,
graph
...
Now run your plot command again and
the file will be saved in the current directory, usually in your My Documents folder (it can also
be found by selecting the “show Current Directory” option in the File menu)
...
png run the following commands:
set terminal png
set output ’graph
...
There are many terminal types (which determine the
output format)
...
In Linux,
the postscript terminal type is popular, since the print quality is high and there are many
PostScript viewers available
...
GNU Free Documentation License
Version 1
...
Preamble
The purpose of this License is to make a manual, textbook, or other functional and useful
document “free” in the sense of freedom: to assure everyone the effective freedom to copy
and redistribute it, with or without modifying it, either commercially or noncommercially
...
This License is a kind of “copyleft”, which means that derivative works of the document
must themselves be free in the same sense
...
We have designed this License in order to use it for manuals for free software, because free
software needs free documentation: a free program should come with manuals providing the
same freedoms that the software does
...
We recommend this License principally for works whose purpose is instruction
or reference
...
APPLICABILITY AND DEFINITIONS
This License applies to any manual or other work, in any medium, that contains a notice
placed by the copyright holder saying it can be distributed under the terms of this License
...
The “Document”, below, refers to any such manual
or work
...
You accept
the license if you copy, modify or distribute the work in a way requiring permission under
copyright law
...
160
GNU Free Documentation License
161
A “Secondary Section” is a named appendix or a front-matter section of the Document
that deals exclusively with the relationship of the publishers or authors of the Document
to the Document’s overall subject (or to related matters) and contains nothing that could
fall directly within that overall subject
...
) The relationship
could be a matter of historical connection with the subject or with related matters, or of
legal, commercial, philosophical, ethical or political position regarding them
...
If a section does not fit the above definition of Secondary then it is not
allowed to be designated as Invariant
...
If the Document does not identify any Invariant Sections then there are none
...
A Front-Cover Text may be at most 5 words, and a Back-Cover Text may be at most
25 words
...
A copy made in an otherwise Transparent file format
whose markup, or absence of markup, has been arranged to thwart or discourage subsequent
modification by readers is not Transparent
...
A copy that is not “Transparent” is called “Opaque”
...
Examples of transparent image formats include PNG, XCF and JPG
...
The “Title Page” means, for a printed book, the title page itself, plus such following pages
as are needed to hold, legibly, the material this License requires to appear in the title page
...
The “publisher” means any person or entity that distributes copies of the Document to
the public
...
(Here XYZ stands for a specific section name mentioned below, such as
“Acknowledgements”, “Dedications”, “Endorsements”, or “History”
...
The Document may include Warranty Disclaimers next to the notice which states that
this License applies to the Document
...
2
...
You may not use technical measures to
obstruct or control the reading or further copying of the copies you make or distribute
...
If you distribute a large enough
number of copies you must also follow the conditions in section 3
...
3
...
Both
covers must also clearly and legibly identify you as the publisher of these copies
...
You may add other material on the covers in addition
...
If the required texts for either cover are too voluminous to fit legibly, you should put the
first ones listed (as many as fit reasonably) on the actual cover, and continue the rest onto
adjacent pages
...
If you use the latter option,
you must take reasonably prudent steps, when you begin distribution of Opaque copies in
GNU Free Documentation License
163
quantity, to ensure that this Transparent copy will remain thus accessible at the stated location until at least one year after the last time you distribute an Opaque copy (directly or
through your agents or retailers) of that edition to the public
...
4
...
In addition,
you must do these things in the Modified Version:
A
...
You may use the same title as a previous version if the
original publisher of that version gives permission
...
List on the Title Page, as authors, one or more persons or entities responsible for authorship of the modifications in the Modified Version, together with at least five of the
principal authors of the Document (all of its principal authors, if it has fewer than five),
unless they release you from this requirement
...
State on the Title page the name of the publisher of the Modified Version, as the publisher
...
Preserve all the copyright notices of the Document
...
Add an appropriate copyright notice for your modifications adjacent to the other copyright notices
...
Include, immediately after the copyright notices, a license notice giving the public permission to use the Modified Version under the terms of this License, in the form shown
in the Addendum below
...
Preserve in that license notice the full lists of Invariant Sections and required Cover
Texts given in the Document’s license notice
...
Include an unaltered copy of this License
...
Preserve the section Entitled “History”, Preserve its Title, and add to it an item stating
at least the title, year, new authors, and publisher of the Modified Version as given on the
Title Page
...
164
GNU Free Documentation License
J
...
These may be placed in the “History” section
...
K
...
L
...
Section numbers or the equivalent are not considered part of the section titles
...
Delete any section Entitled “Endorsements”
...
N
...
O
...
If the Modified Version includes new front-matter sections or appendices that qualify as
Secondary Sections and contain no material copied from the Document, you may at your
option designate some or all of these sections as invariant
...
These titles must be
distinct from any other section titles
...
You may add a passage of up to five words as a Front-Cover Text, and a passage of up to
25 words as a Back-Cover Text, to the end of the list of Cover Texts in the Modified Version
...
If the Document already includes a cover
text for the same cover, previously added by you or by arrangement made by the same entity
you are acting on behalf of, you may not add another; but you may replace the old one, on
explicit permission from the previous publisher that added the old one
...
5
...
The combined work need only contain one copy of this License, and multiple identical Invariant Sections may be replaced with a single copy
...
Make the same adjustment to the section titles
in the list of Invariant Sections in the license notice of the combined work
...
You must delete all
sections Entitled “Endorsements”
...
COLLECTIONS OF DOCUMENTS
You may make a collection consisting of the Document and other documents released under this License, and replace the individual copies of this License in the various documents
with a single copy that is included in the collection, provided that you follow the rules of this
License for verbatim copying of each of the documents in all other respects
...
7
...
When the Document
is included in an aggregate, this License does not apply to the other works in the aggregate
which are not themselves derivative works of the Document
...
Otherwise they must appear on printed covers that bracket the whole aggregate
...
TRANSLATION
Translation is considered a kind of modification, so you may distribute translations of the
Document under the terms of section 4
...
You may include a translation of this License, and all the license notices in the Document,
and any Warranty Disclaimers, provided that you also include the original English version
of this License and the original versions of those notices and disclaimers
...
If a section in the Document is Entitled “Acknowledgements”, “Dedications”, or “History”,
the requirement (section 4) to Preserve its Title (section 1) will typically require changing
the actual title
...
TERMINATION
You may not copy, modify, sublicense, or distribute the Document except as expressly
provided under this License
...
However, if you cease all violation of this License, then your license from a particular
copyright holder is reinstated (a) provisionally, unless and until the copyright holder explicitly and finally terminates your license, and (b) permanently, if the copyright holder fails to
notify you of the violation by some reasonable means prior to 60 days after the cessation
...
Termination of your rights under this section does not terminate the licenses of parties
who have received copies or rights from you under this License
...
10
...
Such new versions will be similar in spirit to the
present version, but may differ in detail to address new problems or concerns
...
gnu
...
Each version of the License is given a distinguishing version number
...
If the Document does not specify a version number of this License, you may choose any
version ever published (not as a draft) by the Free Software Foundation
...
GNU Free Documentation License
167
11
...
A public wiki that anybody can edit is an example of such a server
...
“CC-BY-SA” means the Creative Commons Attribution-Share Alike 3
...
“Incorporate” means to publish or republish a Document, in whole or in part, as part of
another Document
...
The operator of an MMC Site may republish an MMC contained in the site under CC-BYSA on the same site at any time before August 1, 2009, provided the MMC is eligible for
relicensing
...
Permission is granted to copy, distribute and/or
modify this document under the terms of the GNU Free Documentation License, Version 1
...
A copy of the
license is included in the section entitled “GNU Free Documentation License”
...
Texts
...
If you have Invariant Sections without Cover Texts, or some other combination of the
three, merge those two alternatives to suit the situation
...
History
This section contains the revision history of the book
...
1
...
0
Date: 2009-07-01
Author(s): Michael Corral
Title: Trigonometry
Modification(s): Initial version
2
...
1
Date: 2010-03-12
Author(s): Michael Corral
Title: Trigonometry
Modification(s): Made the following changes:
a) Added a section on polar coordinates
b) Changed the format of the page headers
c) Changed the format of the examples
3
...
2
Date: 2012-06-14
Author(s): Michael Corral
Title: Trigonometry
Modification(s): Re-licensed under the terms of the GNU Free Documentation License
version 1
...
168
Index
Symbols
⊥
...
4
A
acute angle
...
38
addition formulas
...
1
altitude
...
41
amplitude
...
119
angle
...
1
central
...
29
general
...
46
initial side of
...
59
negative
...
1
of incidence
...
75
positive
...
29
right
...
1
subtended
...
24
angle of depression
...
14
angular speed
...
59
arc cosecant
...
122
arc cotangent
...
90, 91
arc secant
...
121
arc tangent
...
13
circle
...
95
triangle
...
141
asymptote
...
124
vertical
...
137
average speed
...
25
central angle
...
102
chord
...
3, 6
area
...
87
circumscribed
...
23, 62
segment
...
6
unit
...
100
circumference
...
59
cis
...
11
cofunctions
...
2
complex number
...
144
argument of
...
140
modulus of
...
141
complex plane
...
17
conjugate
...
2
coordinates
...
25
polar
...
146
cosecant
...
106
cosine
...
104
cotangent
...
107
coterminal angles
...
109
D
De Moivre’s Theorem
...
21
distance from earth to sun
...
94
domain
...
78
E
ellipsoid
...
15
equatorial parallax
...
10
Euclid’s formula
...
35
explementary angles
...
75, 77, 85, 86
function
...
102
gradian
...
103
Greek alphabet
...
79
Heron’s formula
...
57
Hipparchus
...
124
hypotenuse
...
65
difference of two angles
...
78
half-angle
...
82
sum of two angles
...
82
imaginary number
...
139
imaginary part
...
139
included angle
...
75
initial side of angle
...
59
inscribed circle
...
23
inscribed triangle
...
18
inverse cosecant
...
122
inverse cotangent
...
126
inverse sine
...
124
inverse trigonometric functions
...
2
L
Law of Cosines
...
38
ambiguous case
...
38, 51
legs of a right triangle
...
9
linear speed
...
49
minute
...
119
modulus
...
52
N
negative angle
...
144
numerical instability
...
133
O
oblique triangle
...
1
obtuse triangle
...
23
odd function
...
120
P
period of a function
...
4
perpendicular bisector
...
116
phase, out of or in
...
146
polar coordinates
...
24
product-to-sum formulas
...
1
Ptolemy’s Theorem
...
139
Pythagorean Theorem
...
6
Q
quadrants
...
87
radius
...
15
radius of sun
...
109, 120
real part
...
146
reference angle
...
32, 34
regular polygon
...
24
Rhind Papyrus
...
1
right triangle
...
32
rpm
...
7
graph of
...
133
second
...
95
segment
...
4
sine
...
104
unit circle definition of
...
79
sinusoidal curves
...
18
slope of a line
...
75
172
Index
solving a triangle
...
25
straight angle
...
95
subtraction formulas
...
82
supplementary angles
...
7
graph of
...
6, 15, 17
terminal side of angle
...
3
triangle
...
38
area
...
10
inscribed
...
2
median of
...
38
obtuse
...
2
trigonometric equations
...
141
trigonometric functions
...
103
U
unit circle
...
17
vertex
...
105
W
worm thread
...
25
Y
y-axis
...
33
Title: Trignometry Full length books
Description: It is a book based on trignometry. It contains full trignometry.
Description: It is a book based on trignometry. It contains full trignometry.