Download that`s trigonometry

THAT’S
TRIGONOMETRY
For information:
Fred W. Duckworth, Jr.
c/o Jewels Educational Services
1560 East Vernon Avenue
Los Angeles, CA 90011-3839
E-mail: [email protected]
Website: www.trinitytutors.com
Copyright © 2008 by Fred Duckworth. All rights reserved. This publication is
copyrighted and may only be copied, distributed or displayed for personal use on an
individual, one-time basis. Transmitting this work in any form by any means, electronic,
mechanical or otherwise, without written permission from the publisher is expressly
prohibited. All copyright notifications must be included and you may not alter them in
any way. Moreover, you may not modify, transform, or build upon this work, nor use this
work for commercial purposes.
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Lesson 1
Why Study Trigonometry?
__________________________________________________________________
Trigonometry – which was originally all about the relationships that exist between the
sides of a right triangle with respect to a given angle – has become an indispensable tool
among architects, surveyors, astronomers, navigators, and others. This is especially true
when it comes to spherical trigonometry, which is concerned specifically with the study
of triangles on the surface of a sphere such as the Earth.
We can use trigonometry to determine distances and heights that cannot easily be
measured directly, such as the height of an unscalable mountain, the altitude of a plane
flying overhead, or the width of a river too deep to ford. Astronomers also use it to
calculate distances to objects in space, and there exist many other much more complex
applications as well.
Most importantly for us, we will need to learn trigonometry in order to understand
calculus in the future. (Calculus is used to calculate the area or volume of shapes or
figures that have curves which change at constantly varying degrees, or to calculate a
moving object’s varying rate of speed at a given point along its path based on the
moment in time it was there.)
__________________________________________________________________
A PRACTICAL APPROACH
Trigonometry uses the techniques that you previously learned from the study of algebra
and geometry. We will introduce and define the trigonometric functions you will need to
learn geometrically rather than in terms of algebraic equations since this is the way they
were originally derived, and we believe virtually all math topics are most easily
understood when presented exactly as they were discovered by their originators.
We will go on to help you develop facility with these functions as well as the ability to
prove basic identities regarding them. This is especially important if you intend to study
calculus, more advanced mathematics, physics and other sciences, or engineering.
One of the things that the state of California wants you to know are the definition of sine
and cosine as y- and x- coordinates of points on a unit circle, which is probably about the
best place to start, so here we go.
3
Lesson 1
The Definitions
We Are Looking for a “Sine”
Trigonometry is a specialized area of geometry that began by quantifying the
relationships—or ratios—between each side of a right triangle. The term is derived
from the Greek word trigonon, meaning triangle, and from the Greek word
metron, meaning measurement.
The six relationships just mentioned are best conceptualized using something called a
unit circle, which is a circle that has a radius of one unit and has its center at the origin of
an x-y rectangular coordinate system or plane. (See the illustration below.)
Inscribed inside of our unit circle we have a unit triangle, which is a right triangle
formed by the x-axis, along with a diagonal line that meets the x-axis at the origin to form
central angle θ (theta), and finally, the top half of a chord that is bisected by the x-axis.
(The bottom half of the chord not pictured. You may recall that a chord is a line segment
that joins two points on a curve.)
4
We call the top half of the chord that is bisected by the x-axis (the bottom half of the
chord is not pictured) sine, a term that is derived from a word meaning “half chord.”
sine
For reasons we will spell out later, we call the other side of our right triangle the cosine.
5
Going back to our unit-circle, let tangent be a line segment parallel to sine and
tangent to the side of the circle.
let tangent be a line segment parallel to cosine and tangent to the top of the circle.
6
The word secant is from the Latin root secare, meaning “to cut.” The name is
fitting for a segment that cuts through the circle.
We will color secant maroon for easy identification.
The cosecant is the line segment that continues to cut through the circle until it unites
with the cotangent.
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Lesson 2
The Trigonometric Functions
__________________________________________________________________
Note that the x-axis, the radius, and sine all form a triangle inside of our unit circle—one
in which the hypotenuse is also the radius. Note also that the side adjacent to angle θ lies
along the x-axis, and that the (x, y) coordinates at the outer end of the hypotenuse identify
the lengths of the sides adjacent to and opposite of angle θ respectively. As already
mentioned, such a triangle is called a unit-triangle.
Our goal now is to “define” the six relationships, or ratios, mentioned earlier, beginning
with sine. To create our first “definition,” we need to find another part of our unittriangle that we can relate to sine in such a way as to end up with a number that is equal
to the length of line segment we have given the name: sine.
We can accomplish this by relating the length of sine to any other side of the triangle that
equals one. So, dividing the length of sine by the hypotenuse will give the length of sine
right back to us.
But, be careful!
Remember that trigonometry was originally all about the relationships that exist between
the sides of a right triangle with respect to a given angle.
So, although we initially labeled sine as the half-chord directly across from angle θ, in
actuality, sine is more accurately described as . . .
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The ratio between the half-chord directly across from angle θ and the hypotenuse
(radius).
Note that the length of “sine” is equal to y, so what we have is …
y
r
So, sine is not really the opposite side, but rather, the relationship between side opposite
and the hypotenuse. And since the half-chord (the line segment we've been calling sine)
is opposite angle θ, in the final analysis what we actually have is . . .
sin  
sin  
y
r
THE TRIGONOMETRIC FUNCTIONS
sin  
y
r
csc  
r
y
cos  
x
r
sec  
r
x
tan  
y
x
cot  
x
y
If you want to memorize all of the algebraic functions as virtually every other
trigonometry course requires you to do, here are links to a couple of sites that will help
you do so.
However, if you want to make memorization unnecessary by being able to literally “see”
what each one of the functions “means,” set aside some time to study the illustrations on
9
the previous pages, spending the time it takes to be able to visualize the relationships
listed below in your “mind’s eye” so that it all comes together making perfectly logical
sense and isn’t just a collection of abstract facts that have absolutely no meaning to you.
REVIEW
The tangent is like the side opposite divided by the side adjacent, which would be
equivalent to sine divided by cosine.
The cotangent is like the side adjacent divided by the side opposite, which is
equivalent to 1 divided by the tangent. That means that cotangent is the reciprocal of
tangent. (It is also like side adjacent divided by side opposite.)
This secant is like the hypotenuse divided by the side adjacent, which is equivalent to
1 divided by cosine. That means that the secant is the reciprocal of cosine.
And finally, the cosecant is like the hypotenuse divided by side opposite, which is the
same as 1 divided by sine. This means that the cosecant is the reciprocal of sine.
As you can see, sine corresponds to the side opposite theta. If we divide the side opposite
by the hypotenuse (which equals one) what we get back is identical to what we started
with: the side opposite. So, now we have our first ratio.
sin  
opposite
hypotenuse
(We abbreviate sine with sin.)
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We can do the same thing in the case of cosine, which corresponds with the side that is
adjacent to angle theta.
cos 
adjacent
hypotenuse
(We abbreviate cosine with cos.)
Now that we have related both the side opposite theta and the side adjacent to theta to our
hypotenuse, it is time to relate them to each other.
The only problem is, to form ratios in which the denominators will be the numeral 1 we
are going to have to use a couple of similar triangles, the first of which is pictured below
(in yellow).
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As you can see, the side adjacent to theta is equal to one (the radius), so by dividing the
side opposite (which corresponds to tangent) by side adjacent (which equals 1) we get
back the same thing we started with, which is side opposite (or tangent).
So we have tan  
opposite
adjacent
To form a ratio that has cotangent as the numerator and 1 as the denominator we are
going to have to use a second similar triangle (once again filled in with yellow).
If we were to manipulate the triangle by making two turns counterclockwise, we would
find that the side that is equal to 1 (the radius) would correspond to side opposite, and
cotangent would correspond to side adjacent, so that we have…
cot  
adjacent
.
opposite
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y
r=1
x
So then, when it comes to defining our ratios, both sine and tangent correspond to side
opposite, while cosine and cotangent correspond to side adjacent.
Now all that remains is relating the hypotenuse (which corresponds to secant and
cosecant) to side opposite and side adjacent.
To divide the hypotenuse of our first similar triangle (which corresponds to secant) by 1
we need to form a ratio with side adjacent, so we have…
sec  
hypotenuse
.
adjacent
13
And to divide the hypotenuse of our second similar triangle (which corresponds to
cosecant) by 1 we need to form a ratio with side opposite, so we have…
csc  
hypotenuse
.
opposite
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The resulting six ratios constitute…
THE TRIGONOMETRIC FUNCTIONS OF RIGHT ANGLES
sin  
opposite
hypotenuse
csc  
hypotenuse
opposite
cos 
adjacent
hypotenuse
sec  
hypotenuse
adjacent
tan  
opposite
adjacent
cot  
adjacent
opposite
However, since the length of side adjacent is equal to x, and the length of side opposite is
equal to y, with the length of the hypotenuse equal to r (radius), we could write the
trigonometric functions of right angles using these three variables instead. However, if
we do, we no longer have the trigonometric functions of right angles. We now have…
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Then again, since side adjacent corresponds to cosine, and side opposite corresponds to
sine, while the hypotenuse corresponds to
Special Angles
45°
You need to know the function values of certain special angles and you really need
to memorize them because you’ll use them so often that deriving them or looking
them up every time would really slow you down.
Let's begin with Functions of 45°.
Look at the 45°-45°-90° unit-triangle illustrated below.
Figure 1.9
Y
r=1
θ
x
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Since the complementary angles are equal, the sides opposite each (sine and
cosine) will also be equal.
By the Pythagorean theorem, sine² + cosine² = (hypotenuse)²
Let a = the length of sine.
Since sine = cosine, a also = the length of cosine
And if hypotenuse = 1, then (hypotenuse)² = 1² = 1
So again, by the Pythagorean theorem we have: a² + a² = 1
2a² = 1
a² = ½
a = √ ½ = 1/√ 2 = (√2 ) / 2
Since a = sin 45° = cos 45°,
sin 45° = cos 45° = (√ 2 ) / 2
Also, b = cos 45° and b = a; therefore
cos 45° = (√2)/2
And tan 45° = side opp / sid adj = cos 45° / sin 45° = 1
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Special Angles
30°60°
18
Lesson 8
Measuring Angles
The state of California expects you to understand the notion of an angle and how to
measure it, in both degrees and radians. You are also expected to be able to convert
between degrees and radians, so let’s start by developing a clear understanding of
degrees.
As you almost assuredly know, a day is measured in units called hours, each of which is
divided into 60 minutes. Similarly, angles are measured in units called degrees, each of
which is also divided into 60 minutes. And just as each minute of time is divided into 60
seconds, each minute of a degree is also divided into 60 seconds.
So, just as a second of time is 601 of a minute…and a minute of time is 601 of an hour, so
it is when measuring degrees of an angle. Traditionally, portions of a degree have been
measured in minutes and seconds. One minute is 601 of a degree and is written like this: 1'.
Moreover, one second is
1
60
of a minute and is written like this: 1".
Consequently, the measure 12° 42' 38" represents 12 degrees, 42 minutes, 38 seconds.
CALCULATING WITH MINUTES AND SECONDS
To add angle measures that not only involve degrees, but minutes and seconds as well,
you have to add the degrees, minutes and seconds separately.
CONVERTING DECIMAL DEGREES
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Also, with the increasing use of calculators, it is now common to measure angles in
decimal degrees. To convert minutes and seconds into decimal degrees, divide the
minutes by 60, and the seconds by 3600.
To convert decimal degrees into minutes and seconds, you obtain the minutes by
multiplying the decimal portion of the measure times 60. (The whole part of the result
will be the number of minutes.) To obtain the number of seconds, multiply the decimal
part of the result times 60 as well.
EXAMPLE:
Convert 34.817° to degrees, minutes, seconds.
Convert 34.817° to degrees, minutes, seconds.
34.817° = 34° + .817°
= 34° + (.817)(60')
= 34° + 49.02'
= 34° + 49' + .02'
= 34° + 49' + (.02)(60")
= 34° + 49' + 1" (rounded)
= 34° 49' + 1"
terminal side
initial side
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The Pythagorean Identities
THE PYTHAGOREAN THEOREM
You know that according to the Pythagorean Theorem a2 + b2 = c2.
Well, returning to our unit triangle, we can see that x2 + y2 = r2
Figure 1.1
Since x2 corresponds to cosine, y2 corresponds to sine, and r2 corresponds to 1, we have
cos2 θ + sin2 θ = 1. Because the Pythagorean theorem is used to get x2 + y2 = r2 this
equation is referred to as a Pythagorean Identity and is written…
sin2 θ + cos2 θ = 1
If we divide both sides by sin2 we get,
sin 2  cos 2 
1


2
2
sin  sin  sin 2 
sin 2 
cos 2 
1
Of course, 2  1 ,
 cot 2  , and
 csc 2  , which means that…
2
2
sin 
sin 
sin 
1 + cot2 θ = csc2 θ
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And finally, if we divide both sides by cos2 we get,
Of course,
sin 2  cos 2 
1


2
2
cos  cos  cos 2 
sin 2 
cos 2 
1
2

tan

,
 1 , and
 sec 2  , which means that…
2
2
2
cos 
cos 
cos 
tan2 θ + 1 = sec2 θ
Thus we have the three Pythagorean Identities.
The state of California expects you to be able to use the Pythagorean theorem to prove
the fact that the identity, cos2 (x) + sin2 (x) = 1 is equivalent, as well as prove the
Pythagorean Theorem as a consequence of this identity, so take the time right now to
make sure that you understand how to perform the necessary calculations on your own,
without any assistance from your notes, the textbook, or any other resource.
You are also expected to prove other trigonometric identities and simplify them by using
the identity cos2 (x) + sin2 (x) = 1.
For example, you may be asked to use this identity to prove that sec2 (x) = tan2 (x) + 1, so
make sure you understand how to perform the necessary calculations to generate the
other two Pythagorean Identities as well.
RECIPROCAL INDENTITIES (PAGE 31)
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Lesson 8
Graphing Functions page 112
The
4.0 Students graph functions of the form f(t) = A sin ( Bt + C ) or f(t) = A cos ( Bt + C)
and interpret A, B, and C in terms of amplitude, frequency, period, and phase shift.
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Lesson 8
Graphing Functions page 112
The
5.0 Students know the definitions of the tangent and cotangent functions and can graph
them.
6.0 Students know the definitions of the secant and cosecant functions and can graph
them.
7.0
MEMORIZE
The tangent of the angle that a line makes with the x-axis is equal to the slope of the line.
(Find out what this means.)
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Lesson 8
Inverse Trigonometric Functions
Page 207
INVERSE TRIGONOMETRIC FUNCTIONS page 207
Function
Domain
Range
y = Sin –1 x
y = Cos –1 x
y = Tan –1 x
y = Cot –1 x
y = Sec –1 x
y = Csc –1 x
–1 ≤ x ≤ 1
–1 ≤ x ≤ 1
Any real number
Any real number
x ≤ –1 or x ≥ 1
x ≤ –1 or x ≥ 1
 /2  y   /2
0 y 
–  /2 < y <  /2
0<y< 
0  y <  /2 or   y < 3  /2 *
0 < y   /2 or  < y  3  /2 *
8.0 Students know the definitions of the inverse trigonometric functions and can graph
the functions.
25
9.0 Students compute, by hand, the values of the trigonometric functions and the inverse
trigonometric functions at various standard points.
Trigonometric Formulas
Addition Formulas for Sines and
Cosines
Page 172
10.0 The state of California expects you to be able to demonstrate an understanding of the
addition formulas for sines and cosines and their proofs and can use those formulas to
prove and/ or simplify other trigonometric identities.
SUMS AND DIFFERENCES OF SINE AND TANGENT
sin( A  B )  sin A cos B  cos A sin B
sin( A  B )  sin A cos B  cos A sin B
tan( A  B) 
tan A  tan B
1  tan A tan B
tan( A  B ) 
tan A  tan B
1  tan A tan B
26
Trigonometric Identities
Half-Angle Identities
Page 183
11.0 Students demonstrate an understanding of half-angle and double-angle formulas for
sines and cosines and use those formulas to prove and/ or simplify other trigonometric
identities.
cos
1  cos A
A

2
2
sin
1  cos A
A

2
2
tan
A
1  cos A

2
1  cos A
tan
A
sin A

2
1  cos A
tan
A 1  cos A

2
sin A
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Trigonometric Identities
Double-Angle Identities
Page 177
11.0 The state of California expects you to be able to demonstrate an understanding of
half-angle and double-angle formulas for sines and cosines and use those formulas to
prove and/ or simplify other trigonometric identities.
cos 2 A  cos 2 A  sin 2 A
cos 2 A  1  2 sin 2 A
cos 2 A  2 cos 2 A  1
sin 2 A  2 sin A  cos A
2 tan A
tan 2 A 
1  tan 2 A
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Lesson 8
Right Triangles
Page 17
12.0 Students use trigonometry to determine unknown sides or angles in right triangles.
29
Triangles and Vectors
The Trigonometric Laws
LAW OF SINES
If any triangle ABC, with side a, b, and c,
a
b

,
sin A sin B
a
c

,
sin A sin C
and
b
c

.
sin B sin C
See Chapter whatever for practice problems using the Law of Sine.
THE LAW OF COSINES
13.0 Students know the law of sines and the law of cosines and apply those laws to solve
problems. PAGE 231 PAGE 243
DETERMINING AREA
14.0 Students determine the area of a triangle, given one angle and the two adjacent sides.
PAGE 234, 245
30
Polar Equations
Polar Coordinates
Pages 283, 284
15.0 Students are familiar with polar coordinates. In particular, they can determine polar
coordinates of a point given in rectangular coordinates and vice versa.
16.0 Students represent equations given in rectangular coordinates in terms of polar
coordinates.
31
Lesson 8
Complex Numbers
There is no real number solution to the equation x2 + 1 = 0.
For such problems, we must turn to a set of numbers that has real numbers as a subset,
that is, the set of complex numbers.
The state of California and not only expects you to be familiar with complex numbers,
but to also be able to represent a complex number in polar form and know how to
multiply complex numbers in their polar form.
OPERATIONS ON COMPLEX NUMBERS
Complex numbers have the form a + bi, where a and b are real numbers and i is the new
number defined by
i  1
or
32
i2 = –1
Each real number is a complex number, since a real number a may be thought of as the
complex number a + 0i. A complex number of the form 0 + bi, where b is nonzero, is
called an imaginary number.
Both the set of real numbers and the set of imaginary numbers are subsets of the set of
complex numbers. However, they share no members in common. Moreover, complex
numbers that are neither imaginary numbers nor real numbers can only be written in
standard form—with two terms and an operation sign, either the plus sign (+) or the
minus sign (–) .
Lesson 8
DeMoivre’s Theorem
The
18.0 Students know DeMoivre's theorem and can give n th roots of a complex number
given in polar form.
If r(cos θ + i sin θ) is a complex number and if n is any real number, then
[r(cos θ + i sin θ)]n = r n(cos nθ + i sin nθ).
33
Lesson 8
Solving Problems Using
Trigonometry
Page 233 Jean Johnson wishes to measure the distance across the Big Muddy River. She
finds that C = 112° 53', A = 31° 06', and b = 347.6 ft. Find the required distance.
34
We can use the law of sines to find a, but first, we must find angle B.
19.0 Students are adept at using trigonometry in a variety of applications and word
problems.
35