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Title: Trigonometry - Brief And It's Formulas
Description: The trigonometric ratios are the ratios between the two sides of a right-angles triangle with respect to an angle and hence they are real numbers. The angle θ taken into consideration may be acute, obtuse or right angle. Various trigonometry formulas related to the basic trigonometric ratios which must be remembered include: 1) sin2 θ + cos2 θ = 1 2) sec2 θ -tan2 θ = 1 3) cosec2 θ - cot2 θ = 1 4) sin θ cosec θ = tan θ cot θ = sec θ cos θ = 1 5) sin2 θ + cos2 θ = 1, so each of them is numerically less than 1. 6) |sin θ| ≤ 1 and |cos θ| ≤ 1 7) -1
Description: The trigonometric ratios are the ratios between the two sides of a right-angles triangle with respect to an angle and hence they are real numbers. The angle θ taken into consideration may be acute, obtuse or right angle. Various trigonometry formulas related to the basic trigonometric ratios which must be remembered include: 1) sin2 θ + cos2 θ = 1 2) sec2 θ -tan2 θ = 1 3) cosec2 θ - cot2 θ = 1 4) sin θ cosec θ = tan θ cot θ = sec θ cos θ = 1 5) sin2 θ + cos2 θ = 1, so each of them is numerically less than 1. 6) |sin θ| ≤ 1 and |cos θ| ≤ 1 7) -1
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Notes from Trigonometry
Steven Butler
c 2001 - 2003
Contents
DISCLAIMER
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Preface
1 The
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usefulness of mathematics
What can I learn from math?
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The ultimate in problem solving
Take a break
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1 What’s special about triangles?
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Pythagorean theorem
The Pythagorean theorem
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Scaling
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Cavalieri’s principle
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1 The wonderful world of π
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2 Circumference and area of a circle
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3 Gradians and degrees
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4 Minutes and seconds
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5 Radian measurement
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6 Converting between radians and degrees
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5 Trigonometry with right triangles
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6 A word of warning
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7 Supplemental problems
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6 Trigonometry with circles
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7 Graphing the trigonometric functions
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8 Why these functions are useful
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9 Supplemental problems
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CONTENTS
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8 Inverse trigonometric functions
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5 So in answer to our quandary
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6 The other inverse trigonometric functions
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7 Using the inverse trigonometric functions
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1 What the equal sign means
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2 Adding fractions
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3 The conju-what? The conjugate
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4 Dealing with square roots
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5 Verifying trigonometric identities
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1 Conditional relationships
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2 Combine and conquer
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3 Use the identities
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4 ‘The’ square root
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5 Squaring both sides
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6 Expanding the inside terms
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Sum formulas for sine and cosine
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Sum and difference formulas for tangent
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1 The area of triangles
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2 The plan
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3 Breaking up is easy to do
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4 The little ones
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5 Rewriting our terms
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6 All together
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7 Heron’s formula, properly stated
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1 Product to sum identities
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1 Our day of liberty
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2 The law of sines
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3 The law of cosines
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4 The triangle inequality
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16 Bubbles and contradiction
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3 A simpler problem
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4 A meeting of lines
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5 Bees and their mathematical ways
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1 One, two, infinity
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2 Limits
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3 The squeezing principle
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4 A limit involving trigonometry
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1 A remarkable formula
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2 Vi`
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1 The wonderful world of vectors
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2 Working with vectors geometrically
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3 Working with vectors algebraically
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4 Finding the magnitude of a vector
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5 Working with direction
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6 Another way to think of direction
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7 Between magnitude-direction and component form
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8 Applications to physics
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9 Supplemental problems
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dot product and its applications
A new way to combine vectors
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Projection
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The perpendicular part
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22 Introduction to complex numbers
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CONTENTS
23 De Moivre’s formula and induction
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5 Enjoying the view
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6 Applying De Moivre’s formula
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7 Finding roots
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These notes may not be distributed in any way in a commercial setting without
the express written consent of the author
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Please report any errors, suggestions or
questions to the author at the following email address
sbutler@math
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edu
vii
Preface
During Fall 2001 I taught trigonometry for the first time
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Over the
course of the next year I taught trigonometry two more times and those notes
grew into the book that you see before you
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These include such topics as
the Pythagorean theorem (Lecture 2), proof by contradiction (Lecture 16), limits
(Lecture 18) and proof by induction (Lecture 23)
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Since these notes grew as a supplement to a textbook, the majority of the
problems in the supplemental problems (of which there are several for almost every
lecture) are more challenging and less routine than would normally be found in a
book of trigonometry (note there are several inexpensive problem books available
for trigonometry to help supplement the text of this book if you find the problems
lacking in number)
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I would like to thank Brigham Young University’s mathematics department for
allowing me the chance to teach the trigonometry class and giving me the freedom
I needed to develop these notes
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The most beautiful proofs and ideas grew out of material that I
learned from him
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We will also outline several broad approaches
to help in developing problem solving skills
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What can I learn from math?
To begin consider the following taken from Abraham Lincoln’s Short Autobiography
(here Lincoln is referring to himself in the third person)
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He began a course of rigid mental discipline with the intent to improve
his faculties, especially his powers of logic and language
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“Euclid” refers to the book The Elements which was written by the Greek
mathematician Euclid and was the standard textbook of geometry for over two
thousand years
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So this raises the question of why he would spend
so much time studying the subject
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Now just as you don’t walk into a gym and start throwing all the weights onto
a single bar, neither would you sit down and expect to solve the most difficult
1
LECTURE 1
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Your ability to solve problems must be developed, and one of the many
ways to develop your problem solving ability is to do mathematics starting with
simple problems and working your way up to the more complicated problems
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When I
played football in high school I would spend just as much time in the weight room
as any member of the team
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Some people have bodies that respond to training and bulk up right away,
and then some bodies do not respond to training as quickly
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Some
people pick up the subject quickly and fly through it, while others struggle to
understand the basics
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Everyone has the ability to understand and enjoy mathematics, be patient, work
problems and practice thinking
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One of the best, and most famous, is How to Solve It by George Polya
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Essentially
there are four steps involved in solving a problem
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Look at the problem
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Clearly
identify these parts
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Mathematicians are not very original and
often use the same ideas over and over, so look for similar problems, i
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, problems
with the same conclusion or the same given information
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Work
through an example
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Check
each step
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THE USEFULNESS OF MATHEMATICS
3
LOOKING BACK—With the problem finished look at the solution
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Can you see your solution at a glance? Can you give a different proof?
You should review this process several times
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Often
times it is just a matter of understanding the problem that prevents its solution
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The ultimate in problem solving
There is one method of problem solving that is so powerful, so universal, so simple
that it will always work
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If it doesn’t
work then try something else
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While this might seem too easy, it is actually a very powerful problem solving
method
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The secret is
to keep trying
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The road to wisdom? Well it’s
plain and simple to express:
err and err and err again
but less and less and less
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4
Take a break
Let us return one last time to the bodybuilding analogy
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The body needs time to heal and grow
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In solving mathematical problems you might sometimes feel like you are pushing
against a brick wall
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In this situation one of the most helpful things to do is to walk away from the
problem for some time
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When you return the problem will often be easier
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First, you have a fresh perspective and you might notice something about
LECTURE 1
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Second, your subconscious mind will often
keep working on the problem and have found a missing step while you were doing
something else
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There is a catch to this
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If you don’t care your mind will stop working on
it
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Lecture 2
Geometric foundations
In this lecture we will introduce some of the basic notation and ideas to be used
in studying triangles
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2
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The first is trigonon which
means “triangle” and the second is metria which means “measure
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Examples of things that we can
measure in a triangle are the lengths of the sides, the angles (which we will talk
about soon), the area of the triangle and so forth
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But there aren’t similar books
dedicated to studying four-sided objects or five-sided objects or so forth
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Imagine that you made a triangle and a square
out of sticks and that the corners were joined by a peg of some sort through a hole,
so essentially the corners were single points
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What happened? The triangle stayed the same and didn’t change its
shape, on the other hand the square quickly lost its “squareness” and turned into
a different shape
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It is this property that makes triangles important
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Returning to our experiment, we can make the square rigid by adding in an
extra side so that the square will be broken up into a collection of triangles, each
of which are rigid, and so the entire square will now become rigid
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GEOMETRIC FOUNDATIONS
6
work with squares and other polygons (many sided objects) by breaking them up
into a collection of triangles
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2
Some definitions on angles
An angle is when two rays (think of a ray as “half” of a line) have their end point
in common
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A picture of an angle is shown below
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The measure of the angle is a number associated with the
angle that tells us how “close” the rays come to each other, another way to think
of the measure of the angle is the amount of rotation it would take to get from one
side to the other
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The most
prevalent is the system of degrees (◦ )
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An angle with measure 180◦
looks like a straight line and is called a linear angle
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Acute angles are angles
that have measure less than 90◦ and obtuse angles are angles that have measure
between 90◦ and 180◦
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acute
right
obtuse
Some angles are associated in pairs
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Two
angles that have their measure adding to 90◦ are called complementary angles
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LECTURE 2
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Similarly, since complementary
angles need to add to 90◦ the complementary angle is 90◦ − 32◦ = 58◦
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3
Symbols in mathematics
When we work with objects in mathematics it is convenient to give them names
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For example if we are in a romantic mood we could use ‘♥’, or any number of
other symbols
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This is because the Greeks were the first to study geometry
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α
β
γ
δ
ζ
η
θ
–
–
–
–
–
–
–
–
alpha
beta
gamma
delta
epsilon
zeta
eta
theta
ι
κ
λ
µ
ν
ξ
o
π
– iota
– kappa
– lambda
– mu
– nu
– xi
– omicron
– pi
ρ
σ
τ
υ
φ
χ
ψ
ω
–
–
–
–
–
–
–
–
rho
sigma
tau
upsilon
phi
chi
psi
omega
Using symbols it is easy to develop and prove relationships
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This is known as the vertical angle theorem
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a
g
b
LECTURE 2
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Similarly, β and γ also form a straight line
and so again we have that β = 180◦ − γ
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One last note on notation
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) to represent points and lowercase letters (a, b, c,
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While it a goal to be consistent it
is not always convenient, however it should be clear from the context what we are
referring to whenever the notation varies
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4
Isoceles triangles
A special group of triangles are the isoceles triangles
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A
useful fact from geometry is that if two sides of the triangle have equal length then
the corresponding angles (i
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the angles opposite the sides) are congruent
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a
b
b
a
The geometrical proof goes like this
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The triangle that is turned over
will exactly match the original triangle and so in particular the angles (which have
now traded places) must also exactly match, i
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, they are congruent
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Combining these two fact
means that in a triangle having equal sides is the same as having equal angles
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Applying the above argument twice shows that
all the angles of such a triangle are congruent
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5
Right triangles
In studying triangles the most important triangles will be the right triangles
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Triangles can be
LECTURE 2
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Namely, right and oblique
...
The right triangles are the easiest to work with and we will eventually study
oblique triangles through combinations of right triangles
...
6
Angle sum in triangles
It would be useful to know if there was a relation that existed between the angles
in a triangle
...
” Is this always true? The answer is, sort of
...
At the North Pole draw two line segments down to the
equator and join these line segments along the equator
...
An example of what this would look like is
shown below
...
)
Now that we have ruined our faith in the sum of the angles in triangles, let us
restore it
...
The sphere behaves differently than
a piece of paper
...
The study of geometric objects on a piece of paper is called
planar geometry or Euclidean geometry, after the greek mathematician Euclid who
first complied the main results of geometry in The Elements
...
e
...
There are other geometries that
are studied that have infinitely many parallel lines going through a point, these
are called hyperbolic geometries
...
LECTURE 2
...
Pictorially, this means that the angles
α and β are congruent in the picture below
...
To see this, start with any triangle
and form two parallel lines, one that goes through one side of the triangle and the
other that runs through the third vertex, such as shown below
...
Notice now that the angles α0 , β 0 and γ form a
linear angle and so in particular we have,
α + β + γ = α0 + β 0 + γ = 180◦
...
Solution We noted earlier that the all of the angles of an equilateral
triangle are congruent
...
2
...
True/False
...
Justify your answer
...
Give a quick sketch of how to prove that if two angles of a triangle are
congruent then the sides opposite the angles have the same length
...
GEOMETRIC FOUNDATIONS
11
b
a
3
...
This is done by
proving a special case, then using the special case to prove the general case
...
2
[You may assume that the area of a rectangle is the base times the height
...
Hint: a right triangle is
half of another familiar shape
...
(What are they? Hint: examples
of each case are shown below, how would you describe the difference
between them?) For each case break up the triangle in terms of right
triangles and use the results from part (i) to show that they also have
the same formula for area
...
A trapezoid is a four sided object with two sides parallel to each other
...
base1
height
base 2
Show that the area of a trapezoid is given by the following formula
...
LECTURE 2
...
In the diagram below prove that the angles α, β and γ satisfy α + β = γ
...
b
g
a
6
...
A triangle can have two obtuse angles
...
(Remember that we are in Euclidean geometry
...
In the diagram below find the measure of the angle α given that AB = AC
and AD = BD = BC (AB = AC means the segment connecting the points
A and B has the same length as the segment connecting the points A and
C, similarly for the other expression)
...
A
D
B
a C
8
...
All triangles are convex
...
convex
non-convex
Show that the angle sum of a quadrilateral is 360◦
...
GEOMETRIC FOUNDATIONS
13
Show that the angle sum of a convex polygon with n sides is (n − 2)180◦
...
True/False
...
Justify your answer
...
3
...
The Pythagorean theorem is correctly stated in the following
way
...
In this theorem, as with every theorem, it is important that we say what our
assumptions are
...
So in particular if you say that the Pythagorean theorem
is a2 + b2 = c2 then you are only partially right
...
Example 1 Use the Pythagorean theorem to find the missing side of
the right triangle shown below
...
THE PYTHAGOREAN THEOREM
15
Solution In this triangle we are given the lengths of the “legs” (i
...
the
sides joining the right angle) and we are missing the hypotenuse, or c
...
616
Note
√ in the example that there are two values given for the missing side
...
In other words it is an expression
that refers to the unique number satisfying the relationship
...
616, is an approximation to the answer (the ‘≈’ sign is used to indicate an
approximation)
...
Make sure when answering the questions that your answer is
in the requested form
...
This seems reasonable,
just taking the square root of each term, but it is not correct
...
This does not work because there are several operations going on in this relationship
...
Rules of algebra dictate which operations must be done first,
for example one rule says that if you are taking a square root of terms being added
together you first must add then take the square root
...
3
...
We will not
try to go through them all but there are books that contain collections of proofs
of the Pythagorean theorem
...
In
dissection we calculate a value in two different ways
...
These two calculations being equal will give birth to relationships,
which if done correctly will be what we are after
...
So starting
with a right triangle we will make four copies and place them as shown below
...
The value that we will calculate is the area of the figure
...
Since the large square has sides of length c
the area of the large square is c2
...
THE PYTHAGOREAN THEOREM
b
c
16
a
a a-b
a-b
a
b
b
c
The second way we will calculate area is in terms of the pieces making up the
large square
...
Each of the triangles has area (1/2)ab and there are four of them
...
1
c2 = (a − b)2 + 4 · ab = (a2 − 2ab + b2 ) + 2ab = a2 + b2
2
3
...
The sketch that you made would
get larger or smaller, but would always appear essentially the same
...
Mathematically, scaling is
when you multiply all distances by a positive number, say k
...
When k < 1 then we are
shrinking distances and everything is getting smaller
...
This is easily seen when the path is a straight line, but it is also
true for paths that are not straight since all paths can be approximated by
straight line segments
...
This is easily seen for rectangles and any other shape can be
approximated by rectangles
...
This is easily seen for cubes and any other shape can be
approximated by cubes
...
THE PYTHAGOREAN THEOREM
17
Example 2 You are boxing up your leftover fruitcake from the holidays
and you find that the box you are using will only fit half of the fruitcake
...
Will the fruitcake exactly fit in the new box?
Solution The new box is a scaled version of the previous box with a
scaling factor of 2
...
In particular the fruitcake
will not fit exactly but only occupy one fourth of the box
...
Scaling plays an important role in trigonometry, though often behind the scenes
...
Two
triangles are similar if the corresponding angle measurements of the two triangles
match up
...
b'
b
a
g
a'
g'
Essentially, similar triangles are triangles that look like each other, but are
different sizes
...
The reason this is important is because it is often hard to work with full size
representations of triangles
...
Such a triangle could never fit inside a
classroom, nevertheless we draw a picture and find a solution
...
Scaling says that the triangle
that is light years across behaves the same way as a similar triangle that we draw
on our paper
...
Solution Since the two triangles are similar they are scaled versions of
each other
...
THE PYTHAGOREAN THEOREM
5
10
7
18
?
only need to multiply the length of 7 by our scaling factor to get our
final answer
...
In order to achieve this we had to scale by a factor
of 2
...
3
...
This is done by cutting our right triangle up into two smaller
right triangles, which are similar as shown below
...
Further, these triangles will have hypotenuses of length a, b and c
...
Similarly, to get from a hypotenuse of length c to a
hypotenuse of length b we would scale by a factor of (b/c)
...
This is because of the
effect that scaling has on areas
...
But these two smaller triangles exactly make up the large triangle
...
So we have,
2
a 2
b
+M
which simplifies to
c 2 = a2 + b 2
...
THE PYTHAGOREAN THEOREM
3
...
Now you push some books to the left and some books to the right to make
a new shape
...
Pictorially,
an example is shown below
...
That is if you take a shape and then
shift portions of it left or right, but never change any of the widths, then the total
area does not change
...
Example 4 Verify Cavalieri’s principle for parallelograms
...
The area of the original shape, the rectangle, is the base times the
height
...
Now looking at the picture below we can make the parallelogram part
of a rectangle with right triangles to fill in the gaps
...
Or in other words,
1
area = (base + a)(height) − 2 · a(height) = (base)(height)
...
THE PYTHAGOREAN THEOREM
3
...
The area of the squares would be a2 , b2 and c2
...
The process is shown below
...
The first step won’t change area because we shifted the squares to parallelograms and using Cavalieri’s principle the areas are the same as the squares we
started with
...
On the final
step we again use Cavalieri’s principle to show that the area is not changed by
shifting from the parallelograms to the rectangles
...
7
The beginning of measurement
The Pythagorean theorem is important because it marked the beginning of the
measurement of distances
...
The way we will think about
distance is as the length of the shortest path that connects the two points
...
In order to use the Pythagorean theorem we need to introduce a right triangle
into the picture
...
The lengths of the legs of the triangle are found by looking at what they
represent
...
The length on the side represents how much we have
changed our y value, which is y1 − y0
...
So we have,
p
distance = (x1 − x0 )2 + (y1 − y0 )2
...
THE PYTHAGOREAN THEOREM
21
(x1 ,y1 )
y1 -y0
(x0 ,y0 )
x1 -x0
Example 5 Find the distance between the point (1
...
2) and the
point (5
...
5)
...
Solution Using the formula just given for distance we have
p
distance = (1
...
7)2 + (4
...
5))2 ≈ 11
...
Use the distance formula to show that the point (x, y) is on a
circle of radius r centered at (h, k) if and only if
(x − h)2 + (y − k)2 = r2
...
Solution The point (x, y) is on the circle if and only if it is distance r
away from the center point (h, k)
...
Squaring the left and right hand sides of the formula we get
(x − h)2 + (y − k)2 = r2
...
This circle is the circle with radius 1 and centered at the origin
...
3
...
Given that the two triangles shown below are similar and the area of the
smaller triangle is 3, then what is the area of the larger triangle?
LECTURE 3
...
Using the Pythagorean theorem find the length of the missing side of the
triangles shown below
...
You have tied a balloon onto a 43 foot string anchored to the ground
...
How high up is the balloon at this time?
4
...
e
...
Show that for any
choice of m and n that (m2 − n2 , 2mn, m2 + n2 ) is a Pythagorean triple
...
Give another proof by dissection of the Pythagorean theorem using the figure
shown below
...
Give yet another proof by dissection of the Pythagorean theorem using
the figure shown below
...
7
...
Two triangles are similar if they have two pairs of angles which
are congruent
...
(Recall that two triangles are similar if
and only if all of their angles are congruent
...
THE PYTHAGOREAN THEOREM
23
a
b
a
c
c
b
8
...
If
you want to still keep a square shape, by what factor should you increase the
length of the sides?
9
...
10
...
Use Cavalieri’s principle to show that the area of a triangle is one half of the
base times the height
...
When Greek historians first traveled to see the great pyramids of Egypt they
ran across a difficult problem, how to measure the height of the pyramids
...
So they
measured the length of the shadow of the pyramid at the same time that
they measured the shadow of a pole of known height and then used scaling
to calculate the height of the pyramid
...
Using this information estimate the height of the
pyramid
...
4
...
It arises in a wide array of mathematical applications, such as
statistics, mechanics, probability, and so forth
...
circumference of a circle
π=
≈ 3
...
diameter of a circle
Since any two circles are scaled versions of each other it does not matter what
circle is used to find an estimate for π
...
And he made a molten sea, ten cubits from the one brim to the other:
it was round all about, and his height was five cubits: and a line of
thirty cubits did compass it round about
...
Using
the definition of π we get
...
24
LECTURE 4
...
2
25
Circumference and area of a circle
From the definition of π we can solve for the circumference of a circle
...
The diameter of a circle is how wide the circle is at its widest point
...
Thus the
diameter which is all the way across is twice the radius which is half-way across
...
The
idea connecting them runs along the following lines
...
Then take these pie shaped wedges and
rearrange them to form a shape that looks like a rectangle with dimensions of half
the circumference and the radius
...
2
Pictorially, this is seen below
...
3
Gradians and degrees
The way we measure angles is somewhat arbitrary and today there are two major
systems of angle measurement, degrees and radians, and one minor system of angle
measurement, gradians
...
Gradians are not very widely used and this
will be our only mention of them
...
ANGLE MEASUREMENT
26
most calculators will have a ‘drg’ button which will convert to and from degrees,
radians and gradians
...
e
...
Degrees splits a full revolution into 360 parts each part being called 1◦
...
Basically, we
don’t know why 360 was chosen, but it does have the nice property of breaking
into smaller pieces easily, i
...
, we can break it into halves, thirds, fourths, fifths,
sixths, eights, tenths and so on and still be working with whole numbers
...
Recall that an angle is composed of two rays that
come together at a point, an angle is in standard position when one of the sides of
the angle, the initial side, is the positive x axis (i
...
to the right of the origin)
...
A negative
number indicates that we move in a clockwise direction
...
With this in mind, we will call two angles
co-terminal if they end up facing the same direction
...
An example of two
angles which are co-terminal are 45◦ and 405◦
...
Example 2 Find an angle between 0◦ and 360◦ that is co-terminal
with the angle 6739◦
...
But with numbers
like this such a process could take quite a while to accomplish
...
7194
...
So in particular our angle makes 18 revolutions plus a little more
...
ANGLE MEASUREMENT
27
to find our angle that we want we can subtract off 18 revolutions and
the result will be an angle between 0◦ and 360◦
...
4
...
For thousands of years the best way to represent a fraction of a number was with fractions
(and sometimes curiously so)
...
To accommodate this they adopted the system of minutes and
seconds
...
One second (denoted by 00 ) correspond to 1/60 of a minute, or 1/3600 of a degree
...
This system of degrees and minutes allowed for accurate measurement
...
As another example, if we let
the equator of the earth correspond to 360◦ then one second would correspond to
about 101 feet
...
For example, Mount Everest is located at approximately
27◦ 590 1600 north latitude and 86◦ 550 4000 west longitude
...
In addition most handheld scientific calculators are also equipped to convert between
the decimal system and D◦ M 0 S 00
...
Example 3 Convert 51
...
Solution It’s easy to see that we will have 51◦ , it is the minutes and
seconds that will pose the greatest challenge to us
...
1265◦ into minutes we multiply by 60
...
1265◦ = 7
...
So we have 70
...
590 to convert
to seconds
...
590 into
seconds we multiply by 60
...
590 = 35
...
Combining
this altogether we have 51
...
400
...
ANGLE MEASUREMENT
4
...
Radian angle measurement can be related to the edge of the unit circle (recall the
unit circle is a circle with radius 1)
...
Similarly as
with degrees a positive angle means you travel counter-clockwise and a negative
angle means you travel clockwise
...
1
q
By scaling, we can also say that the measure of an angle θ is the length of the
arc between the terminal and initial sides divided by the radius of the circle
...
The circumference of the unit circle is 2π and so a full revolution corresponds
to an angle measure of 2π in radians, half of a revolution corresponds to an angle
measure of π radians and so on
...
4
...
In degrees a full revolution corresponds to 360◦ while in radians
a full revolution corresponds to 2π rads
...
This can
be rearranged to give the following useful (if not quite correct) relationship,
180◦
π rads
=1=
...
Example 3 Convert 240◦ to radians and 3π/8 rads to degrees
...
ANGLE MEASUREMENT
29
Solution For the first conversion we want to cancel the degrees and be
left in radians, so we multiply through by (π/180◦ )
...
3
we want to cancel the radians and be left
through by (180◦ /π)
...
5◦
...
7
Wonderful world of radians
If we can use degrees to measure any angle then why would we need any other way
to measure an angle? Put differently, what is useful about radians? The short, and
correct, answer is that when using angles measured in radians a lot of equations
simplify
...
As an example of this simplification, consider the formula for finding the area
of a pie shaped wedge of a circle
...
r
q
To find this area we will use proportions
...
We will do this proportion twice, once
for each system of angle measurement
...
360
area
θ
θr2
=
which
simplifies
to
area
=
...
LECTURE 4
...
8
30
Supplemental problems
1
...
How many paces is it across the pond from
one edge to the other at the widest point? Round your answer to the nearest
pace
...
Suppose that you were to wrap a piece of string around the equator of the
Earth
...
) Hint: you have all of the information you need to answer the
question
...
Suppose that a new system of angle measurement has just been announced
called percentees
...
Using this new system of angle
measurement, complete the following
...
(b) Convert −86
...
(c) Find an angle between 800 and 900 percentees that is co-terminal with
−327 percentees
...
True/False
...
e
...
Justify your answer
...
5
...
Your angle loving friend has brought over a pizza to
share with you and wants to know at what central angle to cut your slice
...
My sister suffers from crustophobia a condition in which she can eat everything on a pizza except for the outer edge by the crust
...
We measure the length of her leftover
crust and find it to be 9 inches
...
e
...
)
LECTURE 4
...
Given that the two circles below are centered at the same point, that the
length of the line segment joining the points A and B is 20 and that the line
segment just touches the inside circle, find the area that is between the two
circles
...
B
A
8
...
m
...
m
...
Lecture 5
Trigonometry with right triangles
In this lecture we will define the trigonometric functions in terms of right triangles
and explore some of the basic relationships that these functions satisfy
...
1
The trigonometric functions
Suppose we take any triangle and take a ratio of two of its sides, if we were to
look at any similar triangle and the corresponding ratio we would always get the
same value
...
Mathematicians would
describe this ratio as an invariant under scaling, that is the ratio is something that
does not change with scaling
...
In particular, we will give names to the ratios of the sides of a right triangle, and
these will be the trigonometric functions
...
In this triangle (as with every triangle) there are three sides
...
e
...
Pictorially, these are located as shown below
...
They are defined in terms of ratios in
32
LECTURE 5
...
opp
tan(θ) =
To help remember these you can use the acronym SOHCAHTOA (pronounced
“sew-ka-toe-a”) to get the relationships for the sine, cosine and tangent function
(i
...
the Sine is the Opposite over the Hypotenuse, the Cosine is the Adjacent over
the Hypotenuse and the Tangent is the Opposite over the Adjacent)
...
3
q
4
Solution First, we can use the Pythagorean theorem to find the length
of the hypotenuse
...
Using the defining ratios we get,
3
,
5
5
csc(θ) =
,
3
sin(θ) =
4
cos(θ) = ,
5
5
sec(θ) = ,
4
3
tan(θ) = ,
4
4
cot(θ) =
...
But there is more
than one way to measure an angle, and 1◦ is not the same angle as 1 rad
...
For example, if you are working a problem that involves
degrees, make sure your calculator is set in degrees and not radians
...
LECTURE 5
...
2
34
Using the trigonometric functions
Now we know how given a right triangle to find the trigonometric functions associated with the acute angles of the triangle
...
If we knew these ratios and the length of one side of the right triangle then we
could find the lengths of the other sides of the right triangle
...
So how do we get these ratios? Historically, the ratios were found by careful
calculations (some of which we will do shortly) for a large number of angles
...
Over the course of the last few decades such books have become obsolete
...
These
calculators can compute quickly and accurately the trigonometric functions, not
to mention being easier to carry around
...
This is shown in the following example
...
At one point in time you notice a rock directly across
from you
...
How wide
is the river? (A badly drawn picture is shown below to help visualize
the situation
...
The angle that we know about is
32◦ and we know that the length of the side adjacent to the angle is
100 feet
...
Looking at our choices for the trigonometric functions we
see that the tangent function relates all three of these, i
...
the angle,
LECTURE 5
...
So we have
opp
tan(32◦ ) =
or
opp = tan(32◦ )100 ft ≈ 62
...
An important step in the above example was determining which trigonometric
function to use
...
To know which
one we look at our triangle
...
We
see how these sides relate to the angle that we know
...
If they had been the
adjacent and hypotenuse sides we would have used the cosine
...
5
...
For example, if we flip the sine function over we get the cosecant
functions
...
tan(θ)
This explains why calculators do not have buttons for all of the trigonometric
functions
...
Another way that we can relate the trigonometric functions is as ratios of
trigonometric functions
...
The two
most important are,
tan(θ) =
sin(θ)
,
cos(θ)
cot(θ) =
cos(θ)
,
sin(θ)
though others also exist
...
4
The Pythagorean identities
The reason that we want to use right ratios of right triangles is becuase we have a
relationship about the sides of a right triangle, namely the Pythagorean theorem
...
TRIGONOMETRY WITH RIGHT TRIANGLES
36
Using the Pythagorean theorem we get the Pythagorean identities, which are
2
2
adj
opp
(adj)2 + (opp)2
2
2
cos (θ) + sin (θ) =
+
=
= 1
...
If we divide both sides by cos2 (θ) and then use the
reciprocal and quotient identities we get,
cos2 (θ) + sin2 (θ)
1
=
2
cos (θ)
cos2 (θ)
1 + tan2 (θ) = sec2 (θ)
...
or
These three equations form the Pythagorean identities and are useful in simplifying expressions
...
5
Trigonometric functions with some familiar
triangles
Up to this point we have done much talking about the trigonometric functions,
but we have yet to find any exact values of the trigonometric functions
...
Later on we will see how to get more exact values from these
...
If we let the lengths of the legs of the right
triangle
be 1 then by the Pythagorean theorem the hypotenuse will have length
√
2
...
2
1
45
1
From this triangle we get the following,
√
√
1
2
1
2
sin(45◦ ) = √ =
,
cos(45◦ ) = √ =
,
2
2
2
2
tan(45◦ ) =
1
= 1
...
TRIGONOMETRY WITH RIGHT TRIANGLES
37
For 30◦ and 60◦ (or π/6 rads and π/3 rads) we start by constructing an equilateral triangle with sides of length 2 and then by cutting it in half we will get
a right triangle with angles of 30◦ and 60◦
...
Using the Pythagorean
theorem we find that the length of the third side
√
of the right triangle is 3
...
2 30
3
60
1
Using this triangle we can find the exact values of the trigonometric functions
for 30◦ ,
√
√
1
3
1
3
◦
◦
◦
sin(30 ) = ,
cos(30 ) =
,
tan(30 ) = √ =
,
2
2
3
3
and for 60◦ ,
√
3
sin(60 ) =
,
2
◦
5
...
1
A word of warning
Sometimes it is tempting to simplify the expression sin(a + b) as sin(a) + sin(b)
...
Erase any thought from your mind of simplifying in this
manner
...
√
sin(30◦ + 30◦ ) = sin(60◦ ) = 3/2
sin(30◦ ) + sin(30◦ ) = 1/2 + 1/2 = 1
These do not match
...
LECTURE 5
...
7
38
Supplemental problems
1
...
Your friend turns to you and asks, “how tall do you think that tower is?”
Looking down at the pamphlet about the tower you only discover that the
tower is leaning 5
...
Frustrated you look
down at your feet and notice that you have no shadow, that is to say that the
sun is directly overhead
...
75 meters
...
How high is the
leaning tower of Pisa? Round your answer to the nearest meter
...
heigh
t
5
...
75
meters
2
...
3
...
4
...
5
...
cos(θ) = sin(90◦ − θ),
cot(θ) = tan(90◦ − θ),
csc(θ) = sec(90◦ − θ)
...
TRIGONOMETRY WITH RIGHT TRIANGLES
a
39
c
b
q
These are known as the complementary angle or cofunction identities
...
e
...
6
...
Hint: use the Pythagorean identity with the cofunction identities proved in
the previous problem
...
Find the exact value for
sin2 (1◦ ) + sin2 (2◦ ) + sin2 (3◦ ) + · · · + sin2 (88◦ ) + sin2 (89◦ )
where the ‘· · · ’ means you continue the pattern
...
Hint: rearrange and use the previous problem
...
Write sin(θ), cos(θ), sec(θ) and csc(θ) as ratios of trigonometric functions
...
One day you happen to find yourself walking along a path and you notice that
by turning 81◦ to your left you see a large building
...
How far
away is the building when you looked the second time?
10
...
The angle of elevation is the angle that you have to look up from the
horizontal, the angle of depression is how much you look down
...
With this in mind, answer the following question
...
To see the top of the tree you have to look up at an angle of elevation of 25◦
...
)
LECTURE 5
...
One day you find yourself walking toward a large mountain
...
After walking
another 5000 feet you now have to look up at an angle of 16◦ to see the top
of the mountain
...
A badly drawn picture is shown below
...
h
12
16
5000 ft
Note: drawing is not to scale
12
...
In the diagram below express b in terms of a and θ
...
q
a
a
b
14
...
Using this complete the following
...
TRIGONOMETRY WITH RIGHT TRIANGLES
41
(a) For a regular polygon with n sides inscribed in the unit circle (an example is shown below on the left for n = 6) express the total length of
the sides of the polygon in terms of n and trigonometric functions
...
(c) There exists a unique number always bigger than the answer to part (a)
and always smaller than the answer to part (b)
...
Round your answers to six decimal places
...
Repeat the previous problem for area in terms of n and trigonometric functions
...
(a) Let the circumference of a polygon denote the sum of the lengths of all
the sides
...
e
...
Hint: the area of
the triangle shown in problem 7 is (b2 cot(θ/2))/4 (you do not have to
prove this, but it is within your reach)
...
What is the number and why? Hint: as n gets large what
does the polygon look like?
Lecture 6
Trigonometry with circles
In this lecture we will generalize the trigonometric functions so that we can use
any angle
...
6
...
Namely, we can only put acute angles in right
triangles (that is angles between 0◦ and 90◦ )
...
To be able to work with the trigonometric functions of any angle we will define
the trigonometric functions by using the unit circle (recall that a unit circle is a
circle with a radius of 1)
...
So for any angle begin by constructing the angle in standard position, that is
the first part of the angles will be the positive x axis
...
This is illustrated below
...
sin(θ)
Note that the values of the trigonometric functions depend only on where the
terminal side of the angle intersects the unit circle
...
TRIGONOMETRY WITH CIRCLES
43
(cos(q ),sin(q ))
1
q
trigonometric functions
...
So we have that every angle is associated with a point on the unit circle, and
every point on the unit circle is associated with an angle (infinitely many of them
actually because every angle has infinitely many co-terminal angles)
...
2
Different, but not that different
We have now defined the trigonometric functions for acute angles in two ways,
namely as ratios in a right triangle and as points on the unit circle
...
So consider the
picture below for any acute angle θ
...
Since this circle is a unit circle the length of the hypotenuse is 1 (i
...
the hypotenuse is the length of the radius)
...
So finding the sine and cosine from ratios we have,
cos(θ) =
x
= x,
1
sin(θ) =
y
= y
...
TRIGONOMETRY WITH CIRCLES
44
Which agrees with finding the sine and cosine as a point on the unit circle
...
Of course,
with the unit circle we can now find the trigonometric functions for any angle
...
3
The quadrants of our lives
When we worked with right triangles the trigonometric functions were always positive
...
To help keep track of the signs we will split the plane up into four parts, this
is done naturally by using the x axis and the y axis
...
Now recall that the sine function corresponds to the y values and so where the y
values are positive the sine function is positive and where the y values are negative
the sine function is negative
...
Once we know the signs of the sine and cosine function we can then find
the sign for the tangent function (recall that the tangent function is a ratio of the
sine and cosine functions)
...
sin +
cos tan sin cos tan +
6
...
For one thing, many real
world problems can be described in terms of right triangles and so it is good to
get an intuitive understanding of the relationships of right triangles
...
TRIGONOMETRY WITH CIRCLES
45
for acute angles
...
This is done by using reference angles, every angle has a reference angle (an
acute angle between 0◦ and 90◦ )
...
From the point where the angle intersects the unit circle drop
a line straight down (or up) to the x axis
...
The acute
angle of this right triangle located at the origin is the reference angle
...
To determine the sign of the angle we note what quadrant the angle lies in and
then use the chart about the signs of the trigonometric functions in the quadrants
to fix any sign problems
...
Solution First we will simplify matters and find a co-terminal angle to
2820◦ between 0◦ and 360◦
...
This is in quadrant IV, and so
the sine function will be negative, the cosine function will be positive
and the tangent function will be negative
...
300
60
In particular we √
have that the reference angle is 60◦
...
Using the
information about the value of the reference angle and the fact that
our angle is in the fourth quadrant we have,
√
√
3
1
◦
sin(2820 ) = −
, cos(2820◦ ) = , tan(2820◦ ) = − 3
...
TRIGONOMETRY WITH CIRCLES
6
...
In
particular, the Pythagorean identities still hold
...
Then recall that the cosine
function is the x value of a point on the unit circle and the sine function is the y
value of a point on the unit circle
...
6
...
Symmetry deals with how an object is
similar to itself
...
The unit circle is highly symmetric (i
...
you can fold it in half any number of
ways and the two halves will overlap)
...
As a first example, consider the two angles θ and −θ
...
e
...
Pictorially, this is seen below
...
Using these
relationships and symmetry we have,
cos(−θ) = cos(θ),
sin(−θ) = − sin(θ)
...
TRIGONOMETRY WITH CIRCLES
47
trigonometric identities
...
tan(−θ) =
sin(−θ)
− sin(θ)
=
= − tan(θ)
cos(−θ)
cos(θ)
Similarly we can show cot(−θ) = − cot(θ), csc(−θ) = − csc(θ) and sec(−θ) =
sec(θ)
...
As a second example of using symmetry, consider the two angles θ and π − θ
(or if you prefer to work in degrees, θ and 180◦ − θ)
...
e
...
Pictorially, this is seen below
...
Using this and the
relationship from symmetry we get,
cos(π − θ) = − cos(θ),
sin(π − θ) = sin(θ)
...
One of the most important ways we use symmetry is finding additional values
for angles that satisfy a given relationship
...
From the second type of symmetry we know that if we go across
horizontally that we do not change the value for the sine function, and in particular
that sin(180◦ − 32◦ ) = sin(32◦ ) = a
...
In general, by using symmetry around the x and y axis we can find additional
values for angles for the cosine and sine functions respectively
...
LECTURE 6
...
7
48
More exact values of the trigonometric functions
We have already been able to get the exact values of the trigonometric functions
for the angles 30◦ , 45◦ and 60◦
...
This will be done by using the unit circle and examining the
points as indicated below
...
Similarly the point that correspond to 90◦ is located at (0, 1) and so
cos(90◦ ) = 0 and sin(90◦ ) = 1
...
Combining this
with what we did last time we have the chart shown below
...
Using these values for the trigonometric functions and reference angles we can
now find the exact value for a large number of angles
...
6
...
So for any (x, y) except the origin
consider the picture shown below
...
TRIGONOMETRY WITH CIRCLES
(x,y)
49
r = x2+y2
q
p
Here r = x2 + y 2 is the distance to the origin and will always be positive
...
y
cos(θ) =
tan(θ) =
This works by taking any point in the plane (x, y) and associating it with (or
scaling it to) a point on the unit circle, namely the point (x/r, y/r), which is
associated with the same angle θ
...
In particular, as with the unit circle, we can associate every point in the plane
except the origin with an angle
...
6
...
Given that the circle shown below is the unit circle match each of the six
trigonometric functions for the angle θ to one of the following lengths, OA,
OB, OC, OD, M C and M D
...
D
M
B
O
q
A
C
LECTURE 6
...
Given that sin(θ) = − 11/6 and that tan(θ) < 0 find the exact values of all
of the trigonometric functions
...
3
...
4
...
5
...
14 57
17
9
24
15
6
...
sec
cot
csc
sec
cot
csc
II I
sec III IV sec
cot
cot
csc
csc
LECTURE 6
...
What symmetry exists between the angle θ and θ + π (or θ + 180◦ if working
in degrees)? Hint: try putting in a couple of values for θ and see how the
two angles compare
...
Repeat
for cos(π + θ)
...
8
...
Using symmetry, write sin((π/2)−θ) in terms of cos(θ) and/or sin(θ)
...
9
...
Here β refers to the reference angle
of θ
...
Verify that if x and
p y are not both 0 then the point (x/r, y/r) is on the unit
circle, where r = x2 + y 2
...
y
sin(θ) = ,
r
r
csc(θ) = ,
y
x
,
r
r
sec(θ) = ,
x
cos(θ) =
y
,
x
x
cot(θ) =
...
TRIGONOMETRY WITH CIRCLES
( )
( )
( )
2p
120 --
135 --
150 --
3p
5p
6
4
-
3
2
-
1 3
- ,
2 2
2
(0,1)
90 --
( )
( )
( )
p
1
2
2
2
,
52
3
,
60 --
2
2
2
2
3 1
,
2 2
,
2
6
225 --
5p
4
3
2
,-
-
240 --
30 --
,-
3
2
2
2
2
4p
1
3
- ,2 2
( )
( )
( )
3
2
2
4
p
6
(1,0)
1
2
p
0 /360 or 0/2p
( )
( )
( )
-
45 --
3 1
,
2 2
(-1,0)
7p
3
2
180 or p
210 --
p
2
1
(0,-1)
270 or
3p
2
The Unit Circle
2
,-
3
2
,-
,-
1
2
330 --
2
315 --
2
300 --
5p
3
7p
4
11p
6
Lecture 7
Graphing the trigonometric
functions
In this lecture we will explore what functions are and develop a way to graphically
represent a function
...
1
What is a function?
We have been taking a lot of time to discuss the trigonometric functions, now let
us step back a second and see what a function is
...
in
out
Function
In our picture of a function there are two openings, one is for what we put into
our function, the other is for what comes out of our function
...
(It is
this uniqueness that determines whether or not the rule, or the machine, is truly
a function
...
For
these functions our input will be angles, and from the last lecture we know that
53
LECTURE 7
...
So for these
two functions the domain is everything
...
)
The range of the function deals with asking the question, what comes out of
our function? For the trigonometric functions our outputs are numbers and in
particular since the sine and cosine functions are based on the unit circle their
values can only be values that are achieved on the unit circle
...
(Mathematically we would denote this by [−1, 1], again this
represents an interval
...
2
Graphically representing a function
When trying to understand the behavior of a function it is often convenient to be
able to look at a representation of the function so that we can see its behavior at
a glance
...
To construct a graph of a function recall that a function works in pairs of
numbers, the input and the corresponding output
...
As an example from last time we showed that sin(0) = 0 and so one point that
is on the curve of the sine function is the point (0, 0)
...
A similar process will create the cosine curve
...
Now let us think
about what happens to our y values as we go around the circle
...
If we now compare this to what happens with the sine graph we
see that it (unsurprisingly) matches up
...
GRAPHING THE TRIGONOMETRIC FUNCTIONS
55
When we deal with graphing the trigonometric functions we will always work in
radians
...
We
will catch a glimpse of these reasons later on
...
3
Over and over and over again
If we were to graph the sine and cosine curves correctly we would have to put in a lot
of values
...
We know that if two angles are co-terminal they will have the same values for
the trigonometric functions, for example sin(x + 2π) = sin(x)
...
To make a graph of the trigonometric function we only need to determine what
it looks like on an interval that contains a complete revolution
...
Functions that have this property are called periodic and the minimum amount
of time it takes to repeat is the period
...
7
...
There are two special types of
symmetry that we will encounter when graphing functions
...
Imagine graphing the function
then folding it in half along the y axis
...
Such a function is called an even function and
satisfies the relationship f (−x) = f (x)
...
The second type of symmetry is around the origin
...
If it looks the
same as before then it is symmetrical around the origin
...
Examples of odd
trigonometric functions are the sine, cosecant, tangent and cotangent functions
...
f (x) = sin2 (x) − cos(x)
LECTURE 7
...
If they are the
same then the function is even, if they are the negative of each other
then the function is odd, if neither of these are true then the function
is neither
...
f (−x) = sin2 (−x) − cos(−x) = (sin(−x))2 − cos(−x)
= (− sin(x))2 − cos(x) = sin2 (x) − cos(x)
= f (x)
So the function is even
...
5
Manipulating the sine curve
Once we have gotten a graph for the sine function (or any function for that matter)
we may want to manipulate the graph
...
These are done by adding constants into our
basic function, i
...
y = a sin(bx − c) + d
...
• The value a
...
e
...
This constant will change the amplitude of the graph, or how
tall the graph is
...
The new amplitude will be the value |a|, that is the absolute value of a
...
If the value of a is negative then in addition to changing the amplitude it
will also flip the curve over
...
This again is outside and so will effect the y values of the graph
...
For an example the curve y = sin(x) + 2 is 2 above the curve y = sin(x), as
seen below
...
GRAPHING THE TRIGONOMETRIC FUNCTIONS
57
2
1
2p
-1
-2
3
2
1
2p
-1
• The value b
...
e
...
This constant will stretch or shrink the graph horizontally
...
For example the function
y = sin(2x) does not have period 2
...
e
...
So in particular the function y = sin(2x) will have period
(2π/2) = π
...
1
-1
p
2p
• The constant c
...
For example the curve y = sin(x − 2) is the graph
y = sin(x) shifted horizontally to the right 2 units, as seen below
...
In particular, it depends on
the value of b
...
LECTURE 7
...
6
58
2p+2
The wild and crazy inside terms
In graphing functions changing the inside terms seems to do things that are
counter-intuitive
...
This function will have period 2π/b and a horizontal shift of c/b
...
To see where these strange values arise recall that one period of the sine curve
corresponds to one revolution around the circle
...
If we are interested in exploring one period of our modified curve
we would do it by finding when the inside expression is 0 (this is the start of the
period) and when it is 2π (this is the end of the period)
...
Note that the start of the period is now at the value c/b, this is why our
horizontal shift is c/b
...
Example 2 Given that the graph shown below is one period of the
sine curve find the amplitude, vertical shift, period and horizontal shift
...
GRAPHING THE TRIGONOMETRIC FUNCTIONS
59
Solution From the graph the height between the lowest and highest
values is 3, and so the amplitude is half that or 3/2
...
The period is the length from the beginning to the end and so is
2π − π/2 = 3π/2
...
With these values in hand we can now start finding a, b c and d
...
The vertical shift is d and so d = 3/2
...
The horizontal shift is
c/b so (c = bπ/2 = 2π/3)
...
2
3
3
2
In this example we had a specific period of the sine curve given to us
...
You can choose any full period to determine your constants
...
Example 3 Given the following function find the amplitude, vertical
shift, period and horizontal shift
...
π
−3
y = 2 sin πx +
3
Answer: From the equation we can read off the amplitude, which is
a = 2 and the vertical shift which is d = −3
...
To find the horizontal shift we take the value of c = −π/3
and divide it by b = π to get a horizontal shift of (−π/3)/π = −1/3
...
We can then use the information about
the amplitude and the period to draw a box that will tightly contain
one period of the curve
...
With the box in place we then draw in one period of the sine
curve, exactly filling the box, to get our required graph
...
LECTURE 7
...
7
5/3
-1/3
-1
-1
-5
-5
60
5/3
Graphs of the other trigonometric functions
We can repeat a similar discussion for the other trigonometric functions but we will
abstain from doing this
...
-3p /2
-p /2
p /2
The tangent curve
3 p /2
-3p /2
-p /2
p /2
3 p /2
The secant curve
These functions have either the sine or cosine in the denominator
...
When this occurs
the function is trying to divide by zero
...
7
...
e
...
There are a
vast array of things that repeat periodically, the rising and setting of the sun, the
motion of a spring up and down, the tides of the ocean and so on
...
GRAPHING THE TRIGONOMETRIC FUNCTIONS
-p
p
The cotangent curve
-p
61
p
The cosecant curve
It turns out that all periodic behavior can be studied through combinations of
the sine and cosine functions
...
We will not go into it at this time as it requires a substantial calculus
background, instead we will look at some pictures to see what can happen as we
combine more and more trigonometric functions
...
These
particular graphs are graphs that arise from looking at the start of the sum
2
2
2
2
2
sin(x) + sin(2x) + sin(3x) + sin(4x) + · · · + sin(nx) + · · ·
1
2
3
4
n
LECTURE 7
...
Can you guess what the function
would look like if we were able to add up all of the terms?
7
...
A test to determine if a graph comes from a function is known as the vertical
line test
...
From the idea
that every input is associated with a unique output explain why the vertical
line test works
...
Determine whether the following functions are even, odd, or neither
...
There is one and only one function that is both even and odd, what function
is it? Justify your answer
...
In the box below draw in one period of the sine curve so that it will completely
fill the box
...
2
1
p
Amplitude:
Horizontal shift:
4p
Period:
Vertical shift:
5
...
They first
present to you the following graph
...
”
LECTURE 7
...
”
Which one is right? Justify your answer
...
Given that y = a sin(bx − c) + d, where a, b, c and d are known values, what
are the largest and smallest values that y can achieve? Justify your answer
...
)
7
...
On the curve mark
the following points: the beginning of the period, the end of the period, the
point where it is maximum and the point where it is minimum
...
This will be done through developing the inverse
trigonometric functions
...
1
Going backwards
Over the last few lectures we have been examining the trigonometric functions
...
Sometimes we might
want to go backwards, that is we have a number and we want to find an angle that
corresponds to it
...
2
Solution From a previous lecture we constructed the table below,
Angle
sin(θ)
cos(θ)
tan(θ)
0◦
0
1
0
30◦
45◦
60◦
90◦
√
√
3/2
1
√1/2 √2/2
2/2 1/2
0
√3/2
√
3/3
1
3 undef
...
64
LECTURE 8
...
2
65
What inverse functions are
An inverse function reverses the direction of our original function
...
Pictorially,
our inverse function machine would look like the following
...
Mathematically, we will say that the inverse function (which we denote as f −1 )
satisfies the following,
f −1 (y) = x
implies
f (x) = y
...
However, we shall soon see that it is not always true that
f −1 (f (x)) = x
...
3
Problems taking the inverse functions
The reason that it is not always true that f −1 (f (x)) = x is because there can be
multiple values of x that map to the same value of y
...
We found that one angle that mapped to the value of 1/2 under the sine function
was 30◦ , but there are other angles
...
This presents a problem because in order to be a function every input must have
a unique output
...
LECTURE 8
...
4
66
Defining the inverse trigonometric functions
To overcome the problem of having multiple angles mapping to the same value we
will restrict our domain before finding the inverse
...
The way we throw out chunks of our curve is somewhat arbitrary
...
It is in this setting that the inverse trigonometric functions
can be created
...
We will create the inverse trigonometric functions for these by first limiting
our domain and then swapping the corresponding inputs and outputs (an inverse
function swaps the role of inputs and outputs and so the domains and ranges also
swap)
...
For example the arcsine function (denoted by ‘arcsin’) is the inverse function of
the sine function and so forth
...
Function Inverse of
arcsin
sin
arccos
cos
arctan
tan
Domain
Range
[−1, 1]
[−π/2, π/2]
[−1, 1]
[0, π]
(−∞, ∞) (−π/2, π/2)
If our inverse functions are returning angles in degrees then the table would be
filled out in the following way
...
Graphically, inverse functions are
the original function (with the domain restricted) which have been flipped across
the line y = x
...
]
Another notation to use for the inverse trigonometric functions is with an exponent of ‘−1’, i
...
arcsin(x) = sin−1 (x)
...
There is the grand
temptation to interpret sin−1 (x) as (1/ sin(x)) = csc(x)
...
INVERSE TRIGONOMETRIC FUNCTIONS
p/2
-1
1
p
-1
67
1
-p/2
The arcsin curve
The arccos curve
p/2
-p/2
The arctan curve
cosecant functions are very different
...
This goes for all of the trigonometric functions
...
5
So in answer to our quandary
So by the nature of the inverse trigonometric functions we have that as long as y
is in the range of the original function that the following equations hold,
sin(arcsin(y)) = y,
cos(arccos(y)) = y,
tan(arctan(y)) = y
...
That is,
when x is between −π/2 and π/2 rads
arcsin(sin(x)) = x
◦
◦
or when x is between −90 and 90 ,
when x is between 0 and π rads
arccos(cos(x)) = x
◦
◦
or when x is between 0 and 180 ,
when x is between −π/2 and π/2 rads
arctan(tan(x)) = x
or when x is between −90◦ and 90◦
...
INVERSE TRIGONOMETRIC FUNCTIONS
8
...
Looking at your calculators you will notice that the calculator has left
these out as well
...
Consider the following
...
We can repeat the procedure for the other two functions and get,
1
1
arcsec(y) = arccos
,
arccot(y) = arctan
...
7
Using the inverse trigonometric functions
Now that we have gone through the work of describing these functions we will use
them in some applications
...
q
3
4
Solution This is a right triangle and so we can represent the trigonometric functions as ratios
...
The trigonometric function that relates these
two sides is the tangent function and so we have,
3
3
≈ 36
...
6435 rads
...
LECTURE 8
...
19◦ or
...
3
This produces only one angle
...
In general there will
be two angles with the same cosine (or same sine) value in a revolution
...
48
...
19
So in addition to 48
...
19◦ or 311
...
19◦ plus multiples of 360◦ and 311
...
In radians our final answer would be,
...
4421 plus multiples of 2π
...
The type of trigonometric expression we will explore is a
trigonometric function composed with an inverse trigonometric function
...
LECTURE 8
...
x−1
cot arccos
x+1
Solution First note that the output of the arccosine function is an angle
and so in particular the term inside represents an angle
...
So we have,
x−1
x−1
θ = arccos
and so
cos(θ) =
...
This process will produce the triangle shown
below
...
Putting it all together we have,
x−1
x−1
cot arccos
= √
...
Set the inside term to an angle θ
...
Use the properties of the inverse trigonometric functions to rewrite the expression as a trigonometric function of θ
...
Using the expression from the previous step draw a right triangle and use
the Pythagorean theorem to find the length of the missing side
...
Using the right triangle from the previous step evaluate the outside trigonometric function
...
INVERSE TRIGONOMETRIC FUNCTIONS
8
...
In order for a function to be invertible for every output there must be a
unique input that gave that value
...
This is known as the horizontal line test
...
True/False
...
Justify your answer
...
Show that for x > 0 that arctan(x) + arctan(1/x) = π/2
...
Show that arccos(x) = 90◦ − arcsin(x) if the arcsine function is returning
angles in degrees or arccos(x) = (π/2) − arcsin(x) if the arcsine function is
returning angles in radians
...
5
...
3
6
...
Given that x > 1, rewrite the following as an algebraic expression
...
Given that x > 0, rewrite the following as an algebraic expression
...
9
...
Namely, it is used to denote identities and conditional relationships
...
We have seen several
examples of this
...
A conditional relationship represents an equation that is sometimes (possibly
never) true
...
For instance in the last lecture
we found that the relationship cos(θ) = 2/3 is satisfied for some but not all θ
...
(Some
mathematical zealots will use the ‘≡’ sign to denote an identity, we shall not
adopt this practice here
...
The most important part of working with identities is being able to
manipulate them, bend them to your will so to speak
...
72
LECTURE 9
...
2
73
Adding fractions
An important skill to have is the ability to add fractions correctly
...
The
process is shown below
...
This is a common mistake,
almost any example shows that this does not work, one example is (1/2) + (1/2) 6=
(1/2) = (1 + 1)/(2 + 2)
...
cos(θ)
sin(θ)
+
1 + cos(θ) sin(θ)
Solution First we will get a common denominator so we can add and
then simplify whatever we have left
...
Often throughout the process of simplifying expressions
and verifying identities we will repeatedly use many of the identities that we have
found up to this point
...
This must be done with exceeding care and can only be done
when both the term on the top and the term on the bottom multiply everything
else
...
As a good practice you should always double check your cancellation
...
LECTURE 9
...
3
74
The conju-what? The conjugate
One very useful algebraic trick to use in simplifying some expressions is the conjugate
...
So for example
the conjugate of 1 + cos(θ) is 1 − cos(θ) (i
...
we changed the sign in the middle)
...
Use of the conjugate is particularly helpful in getting terms that have expressions like 1 ± cos(x) or 1 ± sin(x) in the denominator out of fractional form
...
For example,
(1 − cos(x))(1 + cos(x)) = 1 − cos2 (x) = sin2 (x)
...
1
1 + sin(x)
Solution We will start with the expression and multiply through both
the top and the bottom by the conjugate
...
) Doing this, we get the following
...
This is no
problem when we break up addition in the numerator, i
...
(a+b)/c = (a/c)+(b/c),
but does not work for terms in the denominator
...
WORKING WITH TRIGONOMETRIC IDENTITIES
9
...
When doing so there are some important things to √
remember
...
e
...
e
...
2
√ The second is that the expression x does not always equal x, but rather
2
x = |x|
...
You can drop
the absolute value sign when you are certain that the value will be nonnegative
...
s
1 − cos(θ)
1 + cos(θ)
Solution Note that in the denominator inside the square root that we
have 1 + cos(θ)
...
So starting by multiplying through by the conjugates, we will get the
following
...
But we cannot drop the absolute value sign on the term sin(θ) because it
can sometimes be negative
...
5
Verifying trigonometric identities
Up to this point we have not been verifying identities but just putting tools in place
to simplify expressions
...
When verifying identities the following guidelines are helpful to keep in mind
...
WORKING WITH TRIGONOMETRIC IDENTITIES
76
• Work with one side at a time and manipulate it to the other side
...
It is important not to mix the sides together
...
This is known as circular logic and should be
avoided at all costs
...
Essentially, treat the two sides
as completely separate until you have shown that they are equal
...
So when picking what side to start with
work with the most complicated side and simplify to the other side
...
e
...
• When you are stuck try putting everything in terms of sines and cosines
(this is possible because of the reciprocal and quotient identities)
...
Example 4 Verify the following identity
...
Let us start with the left hand side and put everything
in terms of sine and cosine and see if something marvelous happens
...
WORKING WITH TRIGONOMETRIC IDENTITIES
77
Example 5 Verify the following identity
...
So we will start with the left and try to get the right
...
6
Supplemental problems
1
...
To see why
this does not work show that,
(cos(θ) − sin(θ))2 = (sin(θ) − cos(θ))2 ,
even though,
cos(θ) − sin(θ) 6= sin(θ) − cos(θ)
...
Simplify the following expression so as to remove the square root
...
WORKING WITH TRIGONOMETRIC IDENTITIES
3
...
sec(θ) − tan(θ) sin(θ) = cos(θ)
4
...
sec2 (θ) + csc2 (θ) = sec2 (θ) csc2 (θ)
5
...
sec(θ) + 1
tan(θ)
+
tan(θ)
1 − sec θ
6
...
sin(x)
1 + cos(x)
+
= 2 csc(x)
sin(x)
1 + cos(x)
7
...
sin2 (y)
cos2 (y)
−
= sin(y) = cos(y)
1 + cos(y) 1 + sin(y)
78
Lecture 10
Solving conditional relationships
In this lecture we will work on solving conditional relationships
...
1
Conditional relationships
A conditional relationship is an equation that is sometimes (possibly never) true
...
The main technique that we will develop is taking our relationship and simplifying it down to the point where we have one (or several) equations where we
have a trigonometric function being equal to a number and then using methods
we have learned previously to actually solve for the angle(s)
...
2
Combine and conquer
If your expression has several terms of the same type combine them together
...
)
Example 1 Solve for the acute angle that satisfies the following conditional relationship
...
3 sin(x) − sin(x) = 1 or 2 sin(x) = 1 or
79
sin(x) =
1
2
LECTURE 10
...
Sometimes we will have terms that we cannot combine together, but if we group
all the terms on one side we can factor a common expression out
...
This is by no means surprising, but nonetheless
useful
...
√
2 sin(x) cos(x) = 3 cos(x)
Solution Since we cannot combine these terms let us group them all
together and see what we can factor out
...
We now have two terms multiplying together that give a value of 0 so
our only solutions will be when one of the terms is 0
...
Namely, when does each term
equal 0
...
Since we are dealing with the cosine function on the unit
circle we would go straight down and get our other solution of 270◦ (or
3π/2)
...
We can
look up on a table and see that this will happen at 60◦ (or π/3)
...
Combining these two we will have that our solution is,
60◦ , 90◦ , 120◦ , 270◦
or in radians
π π 2π 3π
, , ,
...
This seems like a quite logical and consistent
LECTURE 10
...
When solving conditional
relationships, we are looking at x over all possibilities and trying to determine
which ones satisfy the relationship
...
So
we have the following rule in solving conditional relationships: You cannot divide
to cancel terms if the term is ever zero
...
3
Use the identities
Sometimes grouping similar terms together and factoring will not be enough, so we
start getting more sophisticated
...
The identities can be used in several ways, primary is their ability
to simplify complex expressions that might be on one side of the equation
...
Example 3 Solve for all the angles for which the following conditional
relationship is satisfied
...
So after staring at the
equation for some time we come up with a plan
...
So let us divide both sides by the term cos(x)
(here it will be alright to divide because we are not cancelling terms)
...
Now we have gotten to our ideal situation, a function being equal to a
number
...
The tangent function is nice because
it is periodic with period of 180◦ or π and so our final solution is,
45◦ + k180◦ for k = 0, ±1, ±2,
...
LECTURE 10
...
4
82
‘The’ square root
When simplifying some expressions we will take the square root
...
For any number that is not 0 there will
always be two square roots
...
This is often denoted by ‘±’
...
10
...
One thing to try is squaring both sides of the formula
...
Before squaring
it might be necessary to move some terms around to get the best results
...
In other words you
will get answers from this process that appear to be a solution but are actually
not
...
(In general it
is a good idea to check your answers when solving any conditional relationship
...
sin(x) − 1 = cos(x)
Solution We seem extraordinarily stuck
...
Now looking at this we want to use the Pythagorean identities in a way
that will simplify this expression
...
So we get the following,
sin2 (x) − 2 sin(x) + 1 = 1 − sin2 (x)
or 2 sin2 (x) − 2 sin(x) = 0
or 2 sin(x)(sin(x) − 1) = 0
...
SOLVING CONDITIONAL RELATIONSHIPS
83
This breaks down into two simpler equations that we can solve
...
We have that sin(x) = 0 at the left-most and right-most points of
the unit circle which are at 0◦ and 180◦ (or 0 and π radians)
...
So now we have our list of answers but more appropriately
we should call these “possible” answers
...
Test the angle
0◦ (0 rads)
90◦ (π/2 rads)
180◦ (π rads)
Which gives
Include?
sin(0) − 1 = cos(0) or -1=1
No
◦
◦
sin(90 ) − 1 = cos(90 ) or 0=0
Yes
sin(180◦ ) − 1 = cos(180◦ ) or -1=-1
Yes
So we had a false solution hanging around
...
10
...
We can expand our methods so that we can
have more interesting expressions inside, such as (4x − 80◦ )
...
Afterwards, we will summarize the
steps that we used
...
1
sin(5x) =
2
Solution First let us look at how the inside expression varies
...
e
...
For a moment let us set θ = 5x
...
This is a much easier problem than the one that we
started with
...
Since
we are using the sine function to get our second value we would go
straight across horizontally and get a second angle of 150◦
...
So we have that,
θ = 30◦ , 150◦ , 390◦ , 510◦ , 750◦ , 870◦ , 1110◦ , 1230◦ , 1470◦ , 1590◦
...
SOLVING CONDITIONAL RELATIONSHIPS
84
Now we recall that θ = 5x
...
So our final
answer is,
x = 6◦ , 30◦ , 78◦ , 102◦ , 150◦ , 174◦ , 222◦ , 246◦ , 294◦ , 318◦
...
Give the inside expression a name such as θ or whatever else is convenient
...
2
...
3
...
10
...
Solve for the unique acute angle θ that satisfies the following,
1
θ = arccos
sin(θ)
...
Find all the solutions between 90◦ and 270◦ to
√
3
◦
...
Find all the solutions between 0 and 2π (or if you prefer to work in degrees
between 0◦ and 360◦ ) to the following conditional relationship,
2 sin(2x) cos(3x) = cos(3x)
...
Find all the solutions between 0 and 2π (or if you prefer to work in degrees
between 0◦ and 360◦ ) to the following conditional relationship,
2 sin(3x) cos(2x) = cos(2x)
...
How many solutions to the equation sin(4x) = 2 exist for x between 0 and
2π
...
6
...
Justify your answer
...
Along the
way we will learn the useful tool of projection
...
1
Projection
A proper discussion of projection must wait until later
...
Namely, given a hypotenuse of a right triangle
and an acute angle we will find expressions for the lengths of the legs of the right
triangle
...
H
q
Using the definition of trigonometric functions as ratios of right triangles we
can find the length of the missing sides
...
So knowing the length of the hypotenuse and an acute angle we can then fill
in the lengths of the other sides of the triangle, as is shown below
...
If we point our flashlight straight down the
85
LECTURE 11
...
e
...
By itself projection may not seem useful, but we can use projection over and
over and over and
...
a b
a
d
g
b
Solution We start with the length a which is a hypotenuse and then
we keep projecting until we reach the side with length b
...
11
...
To do this we will start with right triangles with angles x and y and then combine
them together to form an angle x + y
...
We will throw in one more triangle
to flush out the picture
...
At the same time we will construct another right triangle in the diagram
with an angle of x + y and a hypotenuse of 1 and use projection on that triangle
...
THE SUM AND DIFFERENCE FORMULAS
87
sin(x)sin(y)
x)
sin(x)cos(y)
(
sin
y
1
1
x
cos(x)sin(y)
( x)
s
co
sin(x+y)
x+y
y
cos(x)cos(y)
cos(x+y)
We can use these two diagrams to compute the same lengths in two different
ways
...
Example 2 Use the sum formula for the cosine function to find the
exact value for cos(75◦ )
...
4
11
...
It is usually easier to
modify what we already have
...
THE SUM AND DIFFERENCE FORMULAS
88
So to get our difference formulas we will use our sum formulas and the even/odd’er
identities (i
...
, sin(−x) = − sin(x) and cos(−x) = cos(x))
...
We can use the difference formulas in the same way that we can use the sum
formulas
...
4
Sum and difference formulas for tangent
To find the sum and difference formulas for the tangent function we can combine
the results of the sum and difference formulas for the sine and cosine function
(recall that the tangent function is the sine function over the cosine function)
...
sin(x + y)
sin(x) cos(y) + cos(x) sin(y)
=
cos(x + y)
cos(x) cos(y) − sin(x) sin(y)
cos(x) sin(y)
sin(x) cos(y)
+
cos(x) cos(y)
cos(x) cos(y)
tan(x) + tan(y)
=
=
cos(x) cos(y)
sin(x) sin(y)
1 − tan(x) tan(y)
− cos(x)
cos(x) cos(y)
cos(y)
tan(x + y) =
We can get our difference formula in the same way, but let us instead modify
what we already have
...
1 + tan(x) tan(y)
Example 3 Verify the following,
π
π
tan θ +
· tan θ −
= −1
4
4
LECTURE 11
...
tan(θ) + 1
tan(θ) − 1
·
1 − tan(θ)
1 + tan(θ)
tan(θ) − 1
= −1
=
1 − tan(θ)
π
π
tan θ +
· tan θ −
=
4
4
11
...
In the figure below express the length a in terms of a product of sines and/or
cosines of various angles
...
In the triangle below show that,
c = a cos(β) + b cos(α)
a
b
g
c
b
a
3
...
Nevertheless the relationship still holds for any two angles
...
First observe that the distance between any
two points on a circle is completely determined by the radius of the circle
and the central angle (i
...
, the angle formed by connecting the two points
wth the center of the circle)
...
THE SUM AND DIFFERENCE FORMULAS
90
(i) Explain why the distance between the points (cos(x), sin(x)) and (cos(y), sin(y))
is equal to the distance between the points (cos(x − y), sin(x − y)) and
(1, 0)
...
(iii) Simplify the answer to part (ii) to get the difference formula for cosine
...
4
...
5
...
6
...
cot(y) − cot(x)
Hint: you might want to use a technique similar to what we used for the
tangent function
...
Show that,
cos(x + y) cos(2y) + sin(x + y) sin(2y) =
cos(x + y) cos(2x) + sin(x + y) sin(2x)
...
Using the sum and difference formulas derive the Pythagorean identity, namely
show cos2 (x) + sin2 (x) = 1
...
Given that tan(a) = 1/5, tan(b) = 1/8, tan(c) = 2 and tan(d) = 1, which is
larger, tan(a+b) or tan(c−d)? Justify your answer without using a calculator
...
Given that tan(a) = 2/3 and tan(b) = 1/5 and that a and b are acute angles
show that a + b = π/4
...
11
...
Lecture 12
Heron’s formula
In this lecture we will develop another way to find the area of a triangle
...
12
...
Let us now produce a third
...
One consequence of this is that there is at most
only one triangle with three sides of given lengths
...
So it seems reasonable that knowing only the lengths of the sides of a
triangle that we should be able to find the area of a triangle
...
This formula has been known for quite some time,
Heron himself lived thousands of years ago, and as is the way of mathematics a
large number of proofs have emerged for it
...
Do not worry about reproducing the proof, rather
look for the ideas and connections that are used
...
2
The plan
With any problem that we have we need to start with a plan
...
We already know some ways to find the area of a triangle, so let us try to
91
LECTURE 12
...
We know for instance that the area is half of the base times the height
...
But what if we were to break the triangle up? Let us break up the triangle into
a collection of smaller triangles that we can easily figure out the areas for
...
Then we could just add the areas together and
get it all back in terms of the original lengths of the sides
...
Then add up the areas of these triangles and put everything back in terms of the
lengths of the sides
...
3
Breaking up is easy to do
Start with any triangle, for example, such as the one shown below
...
e
...
This will
produce a picture such as is shown below
...
At the points
LECTURE 12
...
In fact, we will have a total of six right triangles
...
Doing this we get the
picture below
...
4
C
A
A
The little ones
With our picture in place it is quite quick to add up the area of all these little
triangles
...
2
2
2
We would be done except that we do not know what A, B, C and R are
...
e
...
We
need to find a way to rewrite A, B, C and R as expressions of a, b and c
...
5
Rewriting our terms
From the triangle we get the relationships
A + C = b,
A + B = c,
B + C = a
...
But we are here to
learn about trigonometry and so we will jump to the end and get the following,
1
A = (−a + b + c),
2
1
B = (a − b + c),
2
1
C = (a + b − c)
...
HERON’S FORMULA
94
If you feel uneasy about this last step, feel free to check that the calculations are
correct
...
Solving for R is by far the most
interesting step in this proof and we will have to build up to it
...
e
...
Doing so we get,
A
B
C
,
sin(β) = ,
sin(γ) = ,
X
Y
Z
R
R
R
cos(α) = ,
cos(β) = ,
cos(γ) =
...
So
using the sum formulas from last time and substituting in the values we just found
at the appropriate time we will get,
sin(α) =
0 = sin(180◦ ) = sin(α + β + γ) = sin((α + β) + γ)
= sin(α + β) cos(γ) + sin(γ) cos(α + β)
= sin(α) cos(β) cos(γ) + cos(α) sin(β) cos(γ)
+ cos(α) cos(β) sin(γ) − sin(α) sin(β) sin(γ)
ABC
A RR RBR R RC
=
+
+
−
XY Z XY Z XY Z
XY Z
1
(R2 (A + B + C) − ABC)
...
6
or
ABC
R2 =
A+B+C
r
or
R=
ABC
...
First note
that if we add up the outside edge of the big triangle in two ways we get,
1
or A + B + C = (a + b + c) = s,
2
(i
...
we will define this new term s as half the value of the total distance around
the triangle (or half the “circumference” of the triangle))
...
LECTURE 12
...
r
ABC
area = (A + B + C)R = (A + B + C)
A+B+C
p
p
=
(A + B + C)ABC = s(s − a)(s − b)(s − c),
where s = (a + b + c)/2
...
7
Heron’s formula, properly stated
We have now derived Heron’s formula
...
Given a triangle with sides of length a, b and c, then the area enclosed
by the triangle is given by,
p
1
area = s(s − a)(s − b)(s − c) where s = (a + b + c)
...
20
9
17
Solution Since we know the lengths of the sides of this triangle we can
use Heron’s formula to find the total area
...
And so we get,
p
√
area = 23(23 − 9)(23 − 17)(23 − 20) = 5796 ≈ 76
...
12
...
Express cos(x + y + z) in terms of cos(x), cos(y), cos(z), sin(x), sin(y) and
sin(z)
...
2
...
Hint: 3x = x + x + x
...
Show that x = cos(20◦ ) is one solution of the relationship 8x3 − 6x − 1 = 0
...
LECTURE 12
...
Use Heron’s formula to find the area of the triangles shown below
...
12
6
11
15
8
7
5
...
Explain what happened
...
When we have more then one way to compute the same value we can often
combine the two together to derive useful information
...
a
b
g
c
b
a
Use this to find the angle θ in the picture below
...
In particular we
will derive a number of new identities, namely the double angle identity, power
reduction identity and the half angle identity
...
1
Double angle identities
After mathematicians find one relationship they then will go and examine all of
the consequences that come from it
...
One of the easiest consequences of the sum and difference formulas are the
double angle identities
...
And so we get the following,
sin(2x) = sin(x + x)
= sin(x) cos(x) + cos(x) sin(x)
= 2 sin(x) cos(x)
cos(2x) = cos(x + x)
= cos(x) cos(x) − sin(x) sin(x)
= cos2 (x) − sin2 (x)
Example 1 Given that sin(θ) = 3/5 and that cos(θ) = 4/5 find sin(2θ)
and cos(2θ)
...
DOUBLE ANGLE IDENTITY AND SUCH
98
apply the double angle formulas from above and get,
3 4
24
· = ,
5 5
25
2 2
4
3
7
cos(2θ) = cos2 (θ) − sin2 (θ) =
−
=
...
Starting with the double angle identity for the cosine function we can use the
Pythagorean identity to rewrite it in different ways
...
13
...
cos(2x) = 2 cos2 (x) − 1
so
cos(2x) = 1 − 2 sin2 (x)
so
1 + cos(2x)
2
1
−
cos(2x)
sin2 (x) =
2
cos2 (x) =
These are called the power reduction identities since we start with the term on
the left hand side with a square power and the terms on the right side do not have
square power terms in them (i
...
we reduced the highest power term by one)
...
Example 2 Rewrite sin4 (x) to an expression that does not have any
terms with a power greater then one or two different trigonometric
functions multiplied together
...
DOUBLE ANGLE IDENTITY AND SUCH
99
Solution We can use power reduction again and again, doing so we will
get,
1 − cos(2x)
1 − cos(2x)
4
2
2
sin (x) = sin (x) sin (x) =
2
2
2
1 − 2 cos(2x) + cos (2x)
=
4
1 − 2 cos(2x) + ((1 + cos(4x))/2)
=
4
3 − 4 cos(2x) + cos(4x)
=
...
3
Half angle identities
Starting with the power reduction identities we can simultaneously take the square
root of both sides and replace all of the x terms with x/2
...
r
r
x
x
1 + cos(x)
1 − cos(x)
=±
and
sin
=±
cos
2
2
2
2
These equations allow us to find the value of the sine and cosine of half the
angle if we already know the value of the cosine function of the original angle
...
Example 3 Find the exact value of sin(θ/2) and cos(θ/2) given that
cos(θ) = −1/8 and that 3π < θ < 7π/2
...
We already know the range for θ
and so starting with this relationship and dividing through by 2 we get
3π/2 < θ/2 < 7π/4, and so the angle θ/2 lies in the fourth quadrant
...
Now we can proceed, and we get the following,
r
θ
1 − (−1/8)
3
sin
= −
=− ,
2
2
4
r
√
θ
1 + (−1/8)
7
cos
=
=
...
DOUBLE ANGLE IDENTITY AND SUCH
100
To find a half angle identity for the tangent function we can do a similar
procedure, or we can use a combination of the double angle and power reduction
identities
...
2
cos (x/2)
cos2 (x/2)
(1 + cos(x))/2
1 + cos(x)
13
...
In this problem we will compute the exact value of sin(18◦ )
...
(i) Show that if α = 18◦ then sin(2α) = cos(3α)
...
All the
terms have a common factor that can be cancelled
...
(iii) Moving everything over to one side you should now have a quadratic
in terms of sines
...
Now use the
quadratic equation to solve for sin(α), namely you will have,
√
−b ± b2 − 4ac
sin(α) =
...
Hint: you can check your answer by computing sin(18◦ ) and your final answer
with a calculator and checking that they are equal
...
Express sin(4x) in terms of sin(x)’s and cos(x)’s
...
3
...
4
...
π
π
π
π
sin
cos
− cos
sin
9
36
9
36
Hint: this takes more than one step
...
DOUBLE ANGLE IDENTITY AND SUCH
101
5
...
The one most people think
of is the ability to answer questions
...
In mathematics the way that we
look for questions to ask is to look for patterns, things that seem to follow a
behavior
...
(a) As a warmup exercise look at the symbols below
...
To do it you have to identify
the pattern
...
Starting with
cos(π/4) = 2/2 use the half angle identities to find cos(π/8), cos(π/16)
and cos(π/32)
...
(c) Look at the answers for part (b)
...
Looking at that pattern what would you expect the following to be,
π
cos (n+1) = ?
2
√
Note that for n = 1 we have that this is cos(π/4) = 2/2
...
)
6
...
!!
r
x+1
cos 2 arccos
2
(You might want to consider the double angle identity for cosine, namely
cos(2θ) = 2 cos2 (θ) − 1
...
Verify the double angle formula for cotangent, namely,
cot(2x) =
1
1
cot(x) − tan(x)
...
DOUBLE ANGLE IDENTITY AND SUCH
102
8
...
Find the double
angle identity for tangent
...
(b) Given that tan(x) = 2/5 and sin(x) < 0 find the exact value for tan(2x)
...
Given cos(θ) = −7/32 and 5π/2 < θ < 3π find the exact value for sin(θ/4)
...
10
...
Hint: as an intermediate step find cos(θ/2) and sin(θ/2)
...
Show that,
cos(x) + sin(x)
= sec(2x) + tan(2x)
...
Show that,
sec2 (θ) = 2 −
2
...
Write sin(x) cos3 (x) as the sum of sines and/or cosines
...
Rewrite the following as an algebraic expression
...
In particular we will derive the sum to product, product to sum and
identity with no name
...
1
Product to sum identities
The name of this identity tells what it does
...
e
...
e
...
This is possible because the sum
and difference formulas for the cosine function (and similarly for the sine) look
amazingly like each other except for the sign in the middle
...
cos(x + y) + cos(x − y) = (cos(x) cos(y) − sin(x) sin(y)) +
(cos(x) cos(y) + sin(x) sin(y))
= 2 cos(x) cos(y)
If we take this last equation and divide both sides by 2 we get,
1
cos(x) cos(y) = [cos(x + y) + cos(x − y)]
...
2
103
LECTURE 14
...
5◦ ) cos(7
...
Solution We do not have the exact values for either of these angles and
it might take us some time to find them
...
1
[sin(52
...
5◦ ) + sin(52
...
5◦ )]
2
1
=
[sin(60◦ ) + sin(45◦ )]
2√
√
3+ 2
=
4
sin(52
...
5◦ ) =
Example 2 Write the expression 4 cos(3x) cos(5x) as a sum of two
trigonometric functions
...
2
Sum to product identities
These identities do the opposite of the product to sum
...
To derive these identities we will start with the product to sum identities and
use substitution
...
So let us choose to use
new names, say u and v
...
And now for any arbitrary x and y let u = (x + y)/2 and v = (x − y)/2
...
2
2
2
u+v =
LECTURE 14
...
2
2
We can repeat a similar procedure with the other product to sum identities
and get the rest of the sum to product identities
...
2
2
Example 3 Find all of the zeroes between 0◦ and 360◦ to the equation
sin(3x) − sin(x) = 0
...
In particular this is the same as solving,
3x + x
3x − x
2 cos
sin
= 2 cos(2x) sin(x) = 0
...
Since x goes between 0◦ and 360◦ , 2x will go between
0◦ and 720◦ and cosine is zero at the top and bottom of the unit circle,
and so we get,
2x = 90◦ , 270◦ , 450◦ , 630◦
or
x = 45◦ , 135◦ , 225◦ , 315◦
...
So our final answer is,
x = 0◦ , 45◦ , 135◦ , 180◦ , 225◦ , 315◦
...
3
The identity with no name
We will explore one last identity that unfortunately does not have a common name,
but one which is still useful and we will need soon
...
LECTURE 14
...
Let us start with any pair of numbers a and b where at least
one (and usually both) are not zero and consider the point in the plane,
a
b
√
,√
...
But
the points that satisfy x2 +y 2 = 1 are not just any ordinary points, these are points
on the unit circle
...
In particular, there is a unique angle between 0◦ and 360◦ , call it θ, such that,
cos(θ) = √
a
a2 + b2
and
sin(θ) = √
b
...
√
a
b
2
2
a sin(x) + b cos(x) =
a +b √
sin(x) + √
cos(x)
a2 + b 2
a2 + b 2
√
=
a2 + b2 (cos(θ) sin(x) + sin(θ) cos(x))
√
a2 + b2 sin(x + θ)
=
All that remains is to determine the angle θ
...
We could just take the arctangent, but there is a small
technicality, namely the arctangent function only returns angles in a small range
...
In particular, note that there are only two possibilities for θ,
either in the left half or the right half of the plane
...
On the other hand, if it is in the
left half of the plane the arctangent will be off by half of a revolution
...
arctan(b/a)
if a ≥ 0
θ=
arctan(b/a) + 180◦
if a < 0
Example 4 Use the identity with no name to rewrite the following
expression as a single sine function
...
12 sin(3y) − 5 cos(3y)
LECTURE 14
...
All that remains is to find the angle θ
...
62◦ ,
so we can rewrite the expression as,
13 sin(3x − 22
...
14
...
Express sin3 (x) as a sum of sines added together, that is the final answer can
have no sine terms multiplied together
...
Express cos6 (x) as a sum of cosines added together, that is your final answer
can have no cosine terms multiplied together
...
3
...
4
...
5
...
Your final answer should not have any sums or differences
...
6
...
7
...
LECTURE 14
...
Rewrite sin(x) + sin(2x) + sin(3x) + sin(4x) as a product involving trigonometric functions
...
9
...
10
...
cos(x)+cos(y)
11
...
Round your value for θ to two decimal places
...
Given that y = 3 sin(x) − 7 cos(x) find the largest value that y can achieve
...
Hint: if we have y = a sin(t) then the largest
value y can achieve is a and it does so when t = 90◦
...
15
...
We will do it by introducing the law of sines and the law of
cosines
...
For our notation in this lecture we will let a, b and c represent the length of
the sides of a triangle while the quantities α, β and γ will represent the measure
of the corresponding angles
...
a
g
b
b
a
c
15
...
This will split the triangle up into two smaller
right triangles, such as is shown below,
109
LECTURE 15
...
This is done by using projection and we get that,
a sin(β) = h = b sin(α)
a
b
=
...
Combining that result with what we
already have we end up with the following
...
1
a
45
30
Solution We hope to use the law of sines, but before we start we need
to make sure that we can use the law of sines
...
Further, we need to know at least three of
these four (that way we can actually solve for the fourth)
...
So we can use the law of sines and doing so
we get the following,
a
1
=
sin(45◦ )
sin(30◦ )
15
...
sin(30◦ )
The law of cosines
Again we start with our arbitrary triangle and again we draw a line from the vertex
down to the base
...
So
LECTURE 15
...
Since we have right triangles we can use the Pythagorean theorem on these
triangles
...
a2 = (b sin(α))2 + (c − b cos(α))2
= b2 sin2 (α) + c2 − 2bc cos(α) + b2 cos2 (α)
= b2 + c2 − 2bc cos(α)
We can repeat this procedure by dropping down the other vertexes to the other
side and we get some various forms of the same relationship, all of these are various
forms of the law of cosines
...
In
this situation the law of cosines simplifies to give a2 +b2 = c2 , or in other words the
Pythagorean theorem
...
We can also rearrange the terms involved in the law of cosines to solve for the
cosine of the angle
...
Doing this we get the
following equations (also considered forms of the law of cosines)
...
Solution We hope to use the law of cosines
...
LAW OF SINES AND COSINES
4 q
112
6
5
and one angle
...
Since we know the length of all of the sides we
are okay to proceed with using the law of cosines
...
77◦
...
4
The triangle inequality
From the law of cosines we can derive a very important mathematical rule
...
With this in mind, consider the following
...
Triangle inequality
c≤a+b
This also has the alternate forms a ≤ b + c and b ≤ a + c
...
If you want to move from one point on a triangle to another then going on
the segment that connects the two points will always have you travel a distance
that is less than or equal to going along the other two segments
...
If you add up the two shortest sides of a triangle and
it is less than the longest side, then it is no triangle at all
...
Sometimes there is concern over whether this truly is a triangle
...
(Often times by
studying extreme examples, i
...
, worst case scenarios, we can get an idea of some
behavior of an object
...
Particularly in
calculus and any branch of mathematics that has to deal with measurement of
space
...
LAW OF SINES AND COSINES
15
...
Verify the law of sines and the law of cosines if the triangle has an obtuse
angle
...
One day you find yourself walking toward a large mountain
...
After
walking another 5000 feet you now have to look up at an angle of 16◦ to see
the top of the mountain
...
Round your answer to the nearest
100 feet
...
Your friend calls you up and tells you that he is planning to buy a triangular
piece of land with sides of length 4000 feet, 2000 feet, and 6500 feet
...
For our last formula for the area of a triangle, show that the area of the
triangle shown below is given by,
area =
a2 sin(β) sin(γ)
...
LECTURE 15
...
Using the formula from the previous question find the area of the triangles
shown below
...
6
...
Round your answer to two decimal places
...
7
...
Hint: to use the triangle inequality you will need a triangle
...
True/False
...
9
...
Law of tangents
tan (α + β/2)
a+b
=
tan (α − β/2)
a−b
Verify the law of tangents formula
...
LAW OF SINES AND COSINES
115
10
...
Lecture 16
Bubbles and contradiction
In this lecture we will look at an application of what we have learned so far to
show, among other things, why bees make hives with hexagon shapes
...
1
A back door approach to proving
Up to this point we have seen a large amount of mathematics done in a very direct
manner
...
However, in trying to show that a statement is true it is sometimes easier to
take a back door approach to the problem
...
To see how proof by contradiction works imagine that you are shown two envelopes marked A and B
...
You are then told that the envelope marked B does not have the
prize
...
That is the right
action, but let us think carefully why
...
But if we also
knew which one does not have the prize then we can eliminate that as a possibility
and narrow it down to an easier choice, in this case a choice of one
...
But, for any
relationship there are only two possibilities, either it is true or not true
...
We will examine a proof by contradiction
...
116
LECTURE 16
...
2
117
Bubbles
Imagine that you are blowing bubbles
...
Why do bubbles always come in round shapes? It’s because
bubbles form minimum surfaces
...
Bubbles will form a minimum surface because the soap particles that make up
the bubble are attracted to one another
...
This
is why they form spheres
...
If you blow a bunch of bubbles at
one time so that they merge they seem to come together at angles of 120◦
...
This is an interesting property and it would be interesting to try and prove it
...
Instead, we shall
prove a simpler problem
...
3
A simpler problem
In mathematics if we cannot prove something then we turn to the next best thing,
which is to prove a simpler version of what we are trying to do
...
Imagine that you have two pieces of transparent plastic which are connected
by a series of small rods at various points
...
There would now be a collection of soap
films connecting the various rods
...
BUBBLES AND CONTRADICTION
118
From above these soap films would look like a network of lines that connected
the rods
...
We will call any collection of line segments that connects a collection of points
in the plane (connected in the sense that you can go from any point to any other
point by traveling along these line segments) a network
...
Previously we claimed that when bubbles meet that they do so in angles of 120◦
...
Claim: In every soapy network whenever three (or more) lines come
together they will always form angles of 120◦
...
However, this would literally take forever since
there are infinitely many possibilities to check
...
Since a direct approach seems to fail us, let us use our new indirect approach
and use proof by contradiction
...
Let us look at what would make this
statement not true, and call it our anti-claim
...
So to show our claim is true, we will show that our anti-claim is false
...
4
A meeting of lines
So suppose that we have a soapy network connecting a series of points and there
is a time when three or more lines meet and do not form angles all of 120◦
...
If this were not the case then all of the angles would have
to add up to more than 360◦ , but this is impossible
...
This angle is formed by two line segments and we can mark off a small
length on both line segments and get the points B and C (so B and C are both
LECTURE 16
...
So our picture is like
the one shown below
...
)
B
q
A
C
Now with our picture we can draw a line that will bisect the angle θ (i
...
the
line will cut θ in half)
...
If we let the point A0 vary along this angle bisector from the
point A to the midpoint of the segment connecting the points B and C then the
angle that is formed from going from B to A0 to C (denoted by 6 BA0 C) will vary
smoothly from θ to 180◦
...
Now our picture is as shown below
...
These two triangles are congruent (or in other words you
can take one and put it exactly on top of the other)
...
We have,
6
6 A0 AB = 6 A0 AC = θ/2,
AA0 B = 6 AA0 C = 120◦ ,
6 ABA0 = 6 ACA0 = 60◦ − θ/2
...
(Here AB refers to the distance of the line segment connecting
LECTURE 16
...
)
AB
A0 B
A0 A
=
=
sin(120◦ )
sin(θ/2)
sin(60◦ − (θ/2))
We can now solve for the length of the sides of the triangles
...
d sin (60◦ − (θ/2))
θ
d
θ
0
AA=
= d cos
− √ sin
◦
sin(120 )
2
2
3
2d
d sin (θ/2)
θ
= √ sin
A0 B = A0 C =
◦
sin(120 )
2
3
Now consider the following:
A0 B
+
A0 C
+
A0 A
=
=
=
=
=
<
=
2d
θ
θ
2d
√ sin
+ √ sin
2
2
3
3
d
θ
θ
− √ sin
+ d cos
2
2
3
√
θ
θ
d 3 sin
+ d cos
2
2
√
!
3
θ
1
θ
sin
+ cos
2d
2
2
2
2
θ
θ
◦
◦
2d cos(30 ) sin
+ sin(30 ) cos
2
2
θ
2d sin 30◦ +
2
2d
AB + AC
Since 0◦ < θ < 120◦ then 30◦ < 30◦ + θ/2 < 90◦ , it follows that sin(30◦ + θ/2) < 1
...
But we are done
...
In the last step we showed that the total length of the segments A0 B, A0 C and
A0 A is less than the total length of the segments AB and AC
...
We can still connect all of the points together and so the result will still be a
network of the points that we started with
...
BUBBLES AND CONTRADICTION
121
B
A'
A
C
network is now shorter than it was before
...
So our assumption that
we started with must be false
...
In particular, since this is false then the following must be true,
Claim: In every soapy network whenever three (or more) lines come
together they will always form angles of 120◦
...
16
...
It turns out that the hexagon is a shape that can do this
...
Perhaps we underestimate the mathematical
abilities of bees
...
6
Supplemental problems
1
...
2
...
We have not shown that if
LECTURE 16
...
To see this find the total length
of the line segments shown in the two networks below (assume that the angles
formed are all 120◦ , round your answers to two decimal places)
...
e
...
Lecture 17
Solving triangles
In this lecture we will see how to take some information about a triangle and use
it to fill in missing information
...
1
Solving triangles
To solve a triangle means to take some given information about the triangle and
find the measure of the angles and the lengths of the sides of the triangle
...
These are usually some combination of the length of the sides and the measure of
the angles, denoted by S and A respectively
...
For
example if we have a side then an angle then a side (which we shall denote by
SAS) then we will have different information then if we have a side then a side
then an angle (which we shall denote by SSA)
...
a
g
a
b
b
a
SAS
17
...
This is
because if we know two angles of a triangle then we can automatically get the
123
LECTURE 17
...
We can then use the
law of sines to solve for the lengths of the sides
...
e
...
b
a
18
50
33
b
Solution First let us find the angle β
...
So by the law of sines we have,
a
b
18
=
=
...
3
18 sin(33◦ )
≈ 12
...
32
...
Example 2 Fill in the missing information (i
...
solve the triangle)
shown below
...
SOLVING TRIANGLES
125
Solution First we can use the law of cosines to solve for a,
a2 = 102 + 132 − 2(10)(13) cos(73◦ ) ≈ 192
...
89
...
(13
...
4462 so β ≈ 63
...
2(13
...
5◦ = 43
...
17
...
This was not the only way that we could have solved for this angle, we could have
instead used the law of sines
...
This is because the sine function has
some ambiguity built into it
...
e
...
[One thing to note
is that the cosine function does not have this ambiguity and so the law of cosines
will never lead you astray when solving for an angle
...
Given two sides of a triangle the longer side is opposite the bigger angle
...
This will help us correct any ambiguity that we might come across
...
11
b
a
47
6
g
Solution Using the law of cosines we have,
a2 = 62 + 112 − 2(6)(11) cos(47◦ ) ≈ 66
...
18
...
SOLVING TRIANGLES
126
Now the law of sines states,
6
11
8
...
5362 and sin(γ) =
≈
...
8
...
18
Now if we were to plug these into our calculator we would have that
β ≈ 32
...
42◦
...
42◦ + 79
...
84◦
...
What we should state is that after using our calculator
that we have,
β ≈ 32
...
58◦
and
γ ≈ 79
...
58◦
Now the question is which is which
...
Since the side that is opposite the angle β is shorter than the
side that is opposite the angle γ it must be that β is acute
...
42◦
...
58◦
...
If the longest side is not opposite the biggest angle or if
the shortest side is not opposite the smallest angle then somewhere along the lines
there was an error in solving the triangle
...
5
Three sides
A SSS triangle is a triangle where we know the length of all the sides
...
17
...
This is
by far the most interesting possibility
...
SOLVING TRIANGLES
127
one and only one solution, but this has the possibility of no solutions, one solution,
or even two solutions
...
Then start by drawing the
angle α with one side of length b and then the other side extending an unknown
length
...
We then have several possibilities, and these
are shown below
...
)
b
a
b
a
a
a
a
b
a
bsin(a )
a
a=bsin(a )
b
a
a
b=a or b
This is summarized in the following table
...
Notes
0
Side a isn’t long enough
to make a triangle
...
2
Use both the obtuse
and acute angle β
...
When we are actually solving a triangle of this type we will use the law of sines
...
If more then one solution exists, provide both solutions
...
In the previous notation we have that a = 12,
b = 14 and α = 50◦
...
72, and in particular
LECTURE 17
...
So there are two solutions
...
12
14
=
◦
sin(50 )
sin(β)
or
sin(β) =
14 sin(50◦ )
≈
...
34◦ or 116
...
For β = 63
...
34◦ = 66
...
Then
using law of sines we get
12
c
=
◦
sin(66
...
66◦ )
≈ 14
...
sin(50◦ )
For β = 116
...
66◦ = 13
...
Then
using law of sines we get,
c
12
=
◦
sin(13
...
34◦ )
c=
≈ 3
...
sin(50◦ )
Thus our two solutions to our triangle are:
β = 63
...
66◦ ,
17
...
66◦ ,
γ = 13
...
38;
c = 3
...
Surveying
In surveying there are only two things that can be measured directly, namely angles
and distances
...
So essentially surveying boils down to two elements; collecting accurate information about angles and distances, and using that information to solve triangles
...
Example 5 Johnny has been taking his new bride on a tour of the
cow islands
...
He is now planning to return from eight cow
island to the main island on a direct route
...
Solution This can be solved by repeated application of the law of
cosines
...
Namely, we have,
p
distance = 222 + 192 − 2(19)(22) cos(57◦ ) ≈ 19
...
LECTURE 17
...
Namely, we have,
(19
...
18◦
...
74)(19)
We now can find our distance by solving the SAS triangle which consists of going from the Main Island to 4 Cow Island then to 8 Cow
Island
...
74, an angle of 169
...
So we have,
p
Length of trip =
(19
...
74)(16) cos(169
...
6 miles
...
17
...
We have been able to solve triangles given SSS, AAS, ASA, SAA, SAS
and SSA
...
SOLVING TRIANGLES
130
angles of the triangle is it possible to fill in all of the missing values? Justify
your answer
...
There is other information in a triangle then just the measure of the angles
and the lengths of the sides
...
So given that a triangle has angles of 82◦ and 39◦ and an area of 60 square
units, solve the triangle
...
Find all triangles that have sides of length 3 and 5 and an area of 4
...
4
...
Hint:
use the scalene inequality to compare both a and b to a common length
...
Once you have mastered solving triangles you can then solve any shape (i
...
find the missing angles and lengths) by breaking the shape up into triangles,
solving each of those and then putting the triangles back together
...
Round the values for the lengths and the angles
to two decimal places
...
5
9
52
78
7
6
...
Round your answers to
two decimal places
...
7
...
As an example, let points A and B denote the top of two mountains,
we can then measure the angles that are formed by A, B and two points that
LECTURE 17
...
Using the given
information find the distance between A and B (something which is almost
impossible to measure directly)
...
A
B
103
64
49
84
4
...
King Arthur has recently decided to find the speed of the various birds that
he has encountered
...
To
do this he has Polly trained to always fly back to Camelot and then one
day sends Polly out to a nearby village and at noon released, at the same
time ye olde royal hour glass in Camelot starts keeping track of time
...
However, King
Arthur forgot to get an estimate on the distance as a parrot flies between
the village and Camelot
...
Determine the straight line distance and then use this information to determine the airspeed velocity of Polly measured in kilometers per hour
...
LECTURE 17
...
18
...
The problem of how to deal with infinity has plagued mathematicians for thousands
of years
...
He set forth several paradoxes one of which is along the
lines of the following
...
Since
Achilles was 10 times faster than the tortoise (this is a very fast tortoise)
the tortoise was given a 10 meter head start
...
By the time Achilles gets to that point the tortoise has
since moved and now Achilles must get to where the Tortoise is now
at, but again by the time he gets there the tortoise has again moved
...
How then can Achilles win?
The paradox can be easily resolved with the idea that it is possible for infinitely
many things to be added together and get a finite amount (not at all an obvious
fact)
...
By the time that Achilles catches up a second
time the tortoise has moved another
...
By the time that Achilles catches
up a third time the tortoise has moved another
...
And so forth and so
forth
...
INTRODUCTION TO LIMITS
134
If we look at the total distance traveled by Achilles as he is catching up we see
that he will have traveled,
10, 11, 11
...
11, 11
...
1111,
...
e
...
11
...
So when Achilles will have run 100/9 meters he
will then be tied with the tortoise and from there he will easily win the race
...
2
Limits
Often we cannot evaluate something directly, for example, no one has ever actually
added infinitely many numbers together as in the example with Zeno’s paradox,
but we can sometimes say what should happen with great confidence
...
If as
we get closer and closer to what we are trying to evaluate our values are heading
towards a certain number, then we call that number the limit
...
1, 11
...
11
...
It is important to remember that the limit is not necessarily what happens,
only what we expect to happen based on what is happening nearby
...
For example, calculus is essentially the study of two important kinds of
limits (called the derivative and the integral)
...
3
The squeezing principle
We can sometimes find the value of a limit by bounding it above and below in
the limiting process by two other limiting processes which have matching limits at
the place that we are interested in
...
This is called the squeezing principle
...
Now we can only
say that as the plates are coming together that the ball is somewhere in between
the two
...
This is what is happening with the
squeezing principle
...
INTRODUCTION TO LIMITS
18
...
Specifically we are going to determine what happens to the values of the
function,
sin(x)
,
x
as the value of x (measured in radians) approaches 0
...
A
1
x
O B
C
D
From the diagram we have that AB ≤ AC ≤ AD
...
To find AB we can use the right triangle at
the points A, B and the origin which has a hypotenuse of length 1 and an acute
angle x to get that AB = sin(x)
...
To find the
length AC we note that it is an arc of a circle with radius 1 and central angle x
(with x measured in radians), so from geometry we have that AC = x
...
In a fraction if you make a denominator smaller then the total value gets larger
and if you make the denominator larger the total value gets smaller
...
tan(x)
x
sin(x)
We have now been able to put the function sin(x)/x in between the two functions 1 and cos(x) both of which go to 1 as x gets closer and closer to 0
...
Using mathematical notation we would say this in the
following way,
sin(x)
= 1
...
INTRODUCTION TO LIMITS
136
Example 1 Using the relationship,
cos(x) ≤
sin(x)
≤ 1,
x
for x between 0 and π/2 radians show that,
sin(2x)
sin2 (x)
≤
≤ sin(x),
2
x
is also satisfied for x between 0 and π/2 radians
...
Solution First note that sin2 (x)/x = sin(x)(sin(x)/x)
...
Looking at these functions they both go to the value of 0
as x goes to 0
...
18
...
Two trains start out ten miles apart on the same track and head toward
each other, each going at five miles per hour
...
The
fly started on one train and is flying back and forth between the two trains
...
Before the
two trains collide the superfly will have made infinitely many trips back and
forth between the two trains
...
2
...
x→0
x
lim
LECTURE 18
...
(a) Show that for x between 0 and π/2 radians that,
1≤
tan(x)
≤ sec(x)
x
Hint: try to find a way to use what we already know about the expression sin(x)/x between 0 and π/2 radians
...
x
1
0
...
01
1
tan(x)/x
sec(x)
(c) Using the information above and the squeezing principle find the value
for the following
...
(a) Show that for x between 0 and
2 cos2 (x) ≤
π
2
radians that
sin(2x)
≤ 2 cos(x)
...
(b) Using the relationship given in part (a) find
sin(2x)
...
(a) Show that for x between 0 and
0≤
π
2
radians that
1 − cos(x)
sin(x)
≤
...
(b) Using the relationship given in part (a) find
1 − cos(x)
...
INTRODUCTION TO LIMITS
138
6
...
7
...
Show
that the limit of sin(x)/x as x approaches 0 in degrees is π/180
...
19
...
x
x
sin(x) = 2 sin
cos
2
2
x
x
x
= 4 sin
cos
cos
4
2
4
x
x
x
x
= 8 sin
cos
cos
cos
8
2
4
8
= ···
x
x
x
x
n
cos
· · · cos n
= 2 sin n cos
2
2
4
2
If we divide both sides of this equation by x we will have the following
...
To see this let u = (x/2n ), then as n gets large
the value of u goes to zero, since the first term can be simply written as sin(u)/u,
so it follows that the first term is going to 1
...
VIETE’S
FORMULA
140
just keep adding more and more cosine terms on
...
x
x
x
sin(x)
= cos
cos
cos
···
x
2
4
8
Where the ‘· · · ’ mean we keep multiplying cosine terms forever
...
2
Vi`ete’s formula
This last formula is valid for any value of x and so in particular it is valid for the
value x = π/2
...
π
π
π
sin(π/2)
= cos
cos
cos
···
(π/2)
4
8
16
This last expression can be greatly simplified and using a result from an earlier
supplementary problem we get the following relationship which is known as Vi`ete’s
formula
...
However, while it is completely accurate in calculating the value of π it is also extremely slow and so has no practical application
...
20
...
Two examples of this are force and velocity (the magnitude of
velocity is what we commonly call speed)
...
Think of a vector as an object that has both a direction and a magnitude
...
20
...
In the picture below we will connect points A and B with a vector which
−→
we shall denote by AB
...
To help signify
direction we will introduce arrows into our pictures
...
INTRODUCTION TO VECTORS
142
The point that our vector starts at (A in our picture) is called the initial point
and the point that the vector ends at is called the terminal point (B in our picture)
...
Two important operations that can be done with vectors are addition and
scaling
...
For example suppose we want to find
the vector ~u + ~v (i
...
~u and ~v are vectors that we are going to add together), then
put the tail of ~v onto the head of ~u
...
This is shown below
...
If we represented both ways of adding vectors
in the same picture we would see a parallelogram emerge (i
...
the opposite sides
go in the same direction and so are parallel)
...
Vectors turn out to be a good way of
describing relationships of parallelograms
...
Scaling deals
with changing the length of the vector and is done by multiplying a vector by a
constant
...
When the constant
is negative the vector will reverse the direction as well as change length
...
u
2u
-u
(1/2)u
LECTURE 20
...
If we want to find the vector ~u − ~v first start by putting
the tails of the vectors ~u and ~v together
...
This is shown below, also shown below
is the process which shows why this works
...
Multiplying ~v by −1 changed the direction which we represented by reversing the
direction of the arrow of ~v
...
20
...
However, it is often unrealistic to work with vectors in a purely geometric setting
...
So often times we will choose to work with vectors in a form that allows for
more precision in manipulating them
...
Recall that a vector is something that represents a distance and a magnitude
...
Returning to the picture we had at the beginning suppose that A was located
at the point (1, 1) and that B was located at the point (−2, 2)
...
We
−→
will write this as AB = h−3, 1i (we use the ‘h’ and ‘i’ to help distinguish vectors
from points, remember that a vector is not a point but rather a displacement)
...
INTRODUCTION TO VECTORS
144
terminal point B = (x1 , y1 ) then we have,
−→
AB = hx1 − x0 , y1 − y0 i
...
For these operations we will work in each component, and
so we have,
Addition:
Scaling:
ha, bi + hc, di = ha + c, b + di
...
Many of the same rules of arithmetic that we have grown up with still apply
when working with vectors
...
4
=
=
=
=
=
~v + ~u,
~v + (~u + w),
~
~u where ~0 = h0, 0i,
c~u + d~u,
c~u + c~v
...
So let us
look at how to find the magnitude
...
Algebraically, it is not much different
...
If we have a vector ha, bi then we can find
the magnitude by the Pythagorean theorem (see the picture below)
...
LECTURE 20
...
The reason that this works is shown below
...
5
Working with direction
With a way to find magnitude we now turn to direction
...
What exactly is a direction? A good way to think about direction is as
a unit vector
...
A useful fact is that every vector, besides the zero vector (i
...
h0, 0i), can be
represented in a unique way as a positive scalar times a unit vector
...
1
~u = k~uk
~u
k~uk
The important thing to note is that (1/k~uk)~u is a unit vector
...
Example 1 Find a unit vector in the same direction as
h2, −5i
Solution Proceeding with the idea just given we will divide this vector
by its magnitude and get a unit vector pointing in the same direction
...
1
2
1
−5
h2, −5i = √ , √
h2, −5i = p
kh2, −5ik
29 29
22 + (−5)2
There are two very important unit vectors that have been given names, these
are called the standard unit vectors
...
These are
useful in giving another way to represent vectors in component form
...
When you see a vector ~u in the form ai + bj, think of a as how much the vector
is moving in the x direction and b as how much the vector is moving in the y
direction
...
LECTURE 20
...
6
Another way to think of direction
If we have a vector ha, bi that is a unit vector then we also have that a2 + b2 =
kha, bik2 = 1
...
Now we can see trigonometry working its way back into the area of vectors
...
In particular, there is a unique
angle between 0◦ and 360◦ such that (a, b) = (cos(θ), sin(θ))
...
e
...
20
...
So suppose that we are now given a vector, ~v , described as a magnitude
and direction and we want to put the vector into component form
...
We are given an angle θ where we will assume
that θ is measured in standard position
...
Now we have our unit vector, but it may not be the right length
...
So suppose that our magnitude of the vector is k then by
scaling the vector that we just found by k we get the component representation of
the vector, namely,
~v = hk cos(θ), k sin(θ)i = (k cos(θ))i + (k sin(θ))j
...
Suppose that we have a vector that is in component
form and we want to describe it as a magnitude and direction
...
Imagine that the vector has its tail at the origin so that the head of the vector
will be at some point in the plane (a, b) (which by the way are the same a and b as
we use to represent our vector in component form, i
...
the same a and b as ha, bi)
...
INTRODUCTION TO VECTORS
147
Now we recall from when we first learned about the trigonometric functions
that for every point in the plane, besides the origin, there corresponds an angle θ
and further if our point is (a, b) we have,
b
tan(θ) =
...
The
arctangent function only returns values in a 180◦ range, namely between −90◦ and
90◦
...
In particular, we get the following
...
Solution First we find the magnitude
...
7
θ = arctan
≈ 66
...
8◦
...
8
√
58 and is pointing in a direction of
Applications to physics
Many problems in physics reduce down to dealing with vectors
...
If we are using vectors to represent the forces,
then this idea translates into the sum of the vectors being zero
...
9
Supplemental problems
1
...
LECTURE 20
...
Find the exact value for the vector ~a + ~b + ~c where ~a, ~b and ~c are as shown
in the picture above
...
3
...
” Your first event is a three way tug of war in
which three cars are all chained to a central point and try to pull the other
two off
...
The second truck, the
“Rameumpton Raider,” likes to position himself at an angle of 155◦ and pull
with a force of 2100 pounds
...
Hint: for the central point not to move the forces acting on it need to add
up to 0, a badly drawn picture is shown below
...
One day you find yourself out sailing across the ocean
...
The current has
been carrying the boat at a speed of 20 kilometers per hour in a direction of
−25◦ (see the picture below)
...
e
...
If the boat got off the current then
LECTURE 20
...
A badly drawn picture
is shown below
...
Wind
? km/hr, ?
Current and Wind
34 km/hr, -7
Current
20 km/hr, -25
Lecture 21
The dot product and its
applications
In this lecture we will explore a new way of “combining” vectors together, namely
the dot product
...
21
...
This time we will look at a new way of
combining them together, but now the result will not be a vector but rather a
number that we will call the dot product
...
That is we multiply the x components and the y components and we add up the
results
...
For example,
~u · ~v = ~v · ~u,
~u · ~0 = 0,
~u · (~v + w)
~ = ~u · ~v + ~u · w
...
We get
~u · ~u = aa + bb = a2 + b2 = k~uk2
...
THE DOT PRODUCT AND ITS APPLICATIONS
151
In other words the dot √
product can be used to find the magnitude of vectors
...
21
...
One rule
in particular provides an amazing application of the dot product, namely the law
of cosines
...
The third side of the triangle we can represent by the
vector ~u − ~v
...
u
u-v
q
v
The three sides of this triangle have length k~uk, k~v k and k~u − ~v k, so we can
now apply the law of cosines to this triangle and we will get the following
...
This provides an alternative way of finding the dot product and gives rise to
the greatest uses of the dot product
...
~u · ~v
~u · ~v
cos(θ) =
or θ = arccos
...
(The angle between vectors is the angle that is formed by putting the
tails of the vectors together
...
THE DOT PRODUCT AND ITS APPLICATIONS
152
Example 1 Find the dot product of two vectors the first of which has
a magnitude of 12 and a direction of 53◦ and the second of which has a
magnitude of 7 and a direction of 87◦ (where the angles are measured
in standard position)
...
We could of
course find the component form, but let us see if there is not a better
way
...
In particular, the angle between these
two vectors is 87◦ − 53◦ = 34◦
...
~u · ~v = (12)(7) cos(34◦ ) ≈ 69
...
Solution This is a straightforward application of the dot product
...
99◦
365
21
...
In
particular if ~u and ~v are perpendicular we have the following,
~u · ~v = k~ukk~v k cos(90◦ ) = 0
...
In
general, we will say that two vectors whose dot product is zero are orthogonal
...
By this convention we will say that ~0 (i
...
the zero vector) is orthogonal to every
vector
...
THE DOT PRODUCT AND ITS APPLICATIONS
21
...
Now imagine that you had several
people pushing your car as shown in the picture below
...
Ultimately the goal is to move the car forward, but the
people who are pushing on the sides are wasting energy because they are not
pushing exactly in the direction that we want the car to go
...
The part pointing
in the direction of the way we want the car to go is the effective force, i
...
the
amount of force someone pushing from behind would have to give to produce the
same result
...
This process of breaking up a vector into pieces is projection
...
Now take a flashlight and shine it in the direction of the stick with the
flashlight parallel to the ground
...
This is shown below
...
THE DOT PRODUCT AND ITS APPLICATIONS
21
...
The answer to this is a vector and we will find it by determining the direction
and the magnitude of the vector
...
For magnitude, start by imagining that we put the two vectors with their tails
together and then drop a line straight down from the head of ~u to a line that is
extended from ~v to form a right triangle
...
Namely
we have that the magnitude of the projection should be k~uk cos(θ)
...
~v
k~ukk~v k cos(θ)
~u · ~v
proj~v (~u) = (k~uk cos(θ))
=
~v =
~v
k~v k
k~v k2
~v · ~v
21
...
We can use projection to find the part of the vector ~u that lies in the same
direction as ~v
...
e
...
This resulting vector will be orthogonal
to ~v
...
~u · ~v
~u · ~v
~v ) · ~v = ~u · ~v −
~v · ~v = ~u · ~v − ~u · ~v = 0
~v · ~v
~v · ~v
Thus given any vector, ~u, and any other non-zero vector ~v we can break ~u
into two parts, one in the same direction as ~v and one orthogonal to ~v
...
(~u − proj~v (~u)) · ~v = (~u −
LECTURE 21
...
Solution First we will find the parallel part, this is simply projection
and so by using the projection formula we get the following vector,
h3, −4i · h2, 3i
−6
−12 −18
h2, 3i =
h2, 3i =
,
...
13 13
13 13
21
...
If ~u = hcos(θ), sin(θ)i and ~v = hcos(φ), sin(φ)i then find and simplify ~u · ~v
...
)
2
...
3
...
For any two values a and b the vectors ha, bi and hb, −ai are
orthogonal
...
4
...
When the dot product is negative the angle between the vectors
is acute
...
5
...
Round your answer to two decimal places
...
Given a vector ~u with a magnitude of 5 and positioned at an angle of 53◦ ,
find the two vectors which have a magnitude of 8 and so that when you take
the dot product with ~u you get 35
...
LECTURE 21
...
Given that the vectors ~u and ~v are orthogonal show that they satisfy,
k~u + ~v k2 = k~uk2 + k~v k2
...
Given that the vectors ~u and ~v satisfy,
k~u + ~v k2 = k~uk2 + k~v k2 ,
show that ~u and ~v are orthogonal
...
Give any two vectors ~u and ~v , prove the following
...
True/False
...
Justify your
answer
...
(a) Prove the following for any vectors ~x and ~y
...
x
y
y
x
(c) This result is known as the law of parallelograms
...
Show that if k~uk = k~v k that ~u +~v and ~u −~v are orthogonal
...
13
...
If ~u and ~v are two nonzero vectors and the projection of ~u onto
~v is the zero vector then ~u and ~v are orthogonal
...
14
...
Show
that the projection function satisfies the following,
LECTURE 21
...
[Note: any function that satisfies these two properties are linear
...
]
Lecture 22
Introduction to complex numbers
In this lecture we will introduce complex numbers, a number system that includes
i, the imaginary number
...
1
You want me to do what?
Several hundred years ago mathematicians were stuck
...
Unfortunately, they were coming up
short, since if you square a positive number you get a positive number and if you
square a negative number you also get a positive number
...
That is i2 = −1
...
In fact any power of i is equal to one of i, −1, −i, 1, to determine
which one you only need to figure out the remainder of the number when dividing
by 4
...
Example 1 Simplify the expression i765
...
In this form it is now easy
to see what happens
...
INTRODUCTION TO COMPLEX NUMBERS
22
...
Gauss, considered one of the greatest mathematicians of all time, was the first person to show that if we include i (and all that
follows) that we can find the roots of any polynomial
...
A complex number
is a number of the form a + bi where a and b are real numbers (the numbers that
you grew up with and love so well)
...
We will say that two complex numbers are the same if and only if they have
the same real parts and the same imaginary parts
...
3
Working with complex numbers
Just as with real numbers we can add, subtract, multiply and divide complex
numbers
...
This is shown in the following
...
(a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i
Division represents a problem because of the complex number that is in the
denominator
...
To our rescue comes the conjugate
...
So
for example, if we denote conjugation by putting a bar over the number then we
have that a + bi = a − bi
...
a + bi
a + bi c − di
(ac + bd) + (−ad + bc)i
ac + bd
bc − ad
=
·
= 2
=
+ 2
i
c + di
c + di c − di
(c + d2 ) + (−cd + cd)i
c2 + d2
c + d2
Example 2 Simplify the following complex number
...
INTRODUCTION TO COMPLEX NUMBERS
160
Solution This is a matter of careful manipulation
...
8 + 15i
27
(8 + 15i) (7 + i)
27
(5 − 5i)
−
=
·
−
·
7−i
5 + 5i
(7 − i) (7 + i) 5 + 5i (5 − 5i)
56 + 8i + 105i − 15 135 − 135i
=
−
72 + 12
52 + 52
−94 + 248i
=
= −1
...
96i
50
22
...
We would like an analogous system for the complex numbers
...
e
...
So geometrically instead of a line we will use a plane
where every point corresponds to a number and every number corresponds to a
point
...
In particular, we will associate the x coordinate of the point with the real part
of the complex number and the y coordinate of the point with the√imaginary part
of the complex number
...
22
...
First, let us consider what the absolute
value function does for real numbers
...
e
...
-3
0
2
We know that |2| = 2 and | − 3| = 3 but what does that mean
...
In particular, the absolute value function measures distance away
from zero
...
INTRODUCTION TO COMPLEX NUMBERS
161
For working with absolute value in the complex numbers (which is sometimes
referred to as modulus) we will adopt this same idea
...
To find this distance think of the geometrical picture
...
So by the Pythagorean theorem
we get,
√
|a + bi| = a2 + b2
...
We can also use it to measure the distance between any two arbitrary complex
numbers
...
(In terms of the picture think of the “−z2 ”
as putting the origin at z2 and then it becomes just a matter of finding the distance
from a number to the origin
...
Solution This is a straightforward application of applying the distance
formula and so we will get
...
6
Trigonometric representation of complex numbers
As with vectors we can describe complex numbers in more than one way
...
In terms of the graphical representation r and θ are related to a complex number
a + bi as shown below
...
e
...
So given a
complex number z = a + bi we have,
√
r = |z| = a2 + b2
...
INTRODUCTION TO COMPLEX NUMBERS
162
a+bi
r
q
To find the angle θ we use the fact that every point except the origin is associated with an angle in the plane
...
b
tan(θ) =
...
To fix this we
will find θ in one of two ways depending on where a + bi is in the complex plane
...
arctan(b/a)
if a ≥ 0
θ=
◦
arctan(b/a) + 180
if a < 0
If we wanted our angle in radians we would replace 180◦ by π
...
In particular if we add any multiple
of a full revolution to θ then we will still get the same complex number (θ + 360◦
or θ + 2π for example)
...
22
...
Suppose we have z1 = r1 (cos(θ1 )+i sin(θ1 ))
and z2 = r2 (cos(θ2 ) + i sin(θ2 )) then for multiplication we get the following
...
LECTURE 22
...
z1
r1
= (cos(θ1 − θ2 ) + i sin(θ1 − θ2 ))
z2
r2
22
...
Find the exact value of
i + i2 + i3 + i4 + i5 + · · · + i7658 + i7659
where the ‘· · · ’ means that you continue the pattern, that is add up all of
the powers of i from 1 to 7659
...
2
...
Hint: recall that z refers to the conjugate of z, one way
to start is to write z = a + bi and then work through the calculations
...
What is the distance between 4 − i and −2 + 3i?
4
...
A picture is an acceptable answer
...
In particular, we will
prove De Moivre’s formula which will allow us to easily find powers and roots of
complex numbers
...
1
You too can learn to climb a ladder
Mathematical induction is a powerful tool that allows us to prove an infinite number of cases of a problem very quickly
...
Namely, proving the
first case and then proving that if one case is true then it must be that the next
case is also true
...
To climb any ladder you first have to get on the ladder and then you have to move
from rung to rung to get up
...
Conversely, if you have not yet learned
to climb a ladder then studying mathematical induction will help teach you how
...
2
Before we begin our ladder climbing
Before climbing up a ladder we would place it against a wall or some other solid
surface, that is we would prepare
...
Before we begin we must understand what we are
trying to prove
...
DE MOIVRE’S FORMULA AND INDUCTION
165
The best way to learn mathematics is to do mathematics and so as we are
learning about induction we will do an induction problem
...
De Moivre’s formula is stated in the following way,
(cos(θ) + i sin(θ))n = cos(nθ) + i sin(nθ)
for n = 1, 2, 3,
...
It basically gives an easy way to find powers of a
particular type of complex number
...
Could you imagine having to actually take that term and multiply it 360 times,
this is going to be such a cool formula
...
23
...
They first have to get on the ladder
...
After all, we are trying to show that infinitely many cases
are true and so we should at least show the first one is true
...
We are trying to show De Moivre’s formula for n = 1, 2, 3,
...
So to prove the first case we put n = 1 into both
sides of De Moivre’s formula and verify that they are equal
...
We have one case down,
now infinitely many to go
...
DE MOIVRE’S FORMULA AND INDUCTION
23
...
For example the process of moving from the first to
the second rung is no different then moving from the 121st to the 122nd rung
...
For example, to
get from one rung to the next you would first move one foot up and then the other
(but not both at the same time)
...
So think of the rungs of the ladder as the individual cases that we are
trying to prove
...
e
...
e
...
To do this we will start by assuming that the kth case is true and then prove
that the (k + 1)st case must also be true
...
This is the most challenging part of an induction proof
...
Shown below is the argument for this step for De Moivre’s
LECTURE 23
...
Look carefully for how we use information about the kth case to simplify
...
23
...
Once we are on
the first step we can get to the second step
...
And so on and so on until we run out of steps
...
Mathematical induction follows the same way
...
We also know that since the first case is true that
the second case must also be true (this is by the second step)
...
And so on and so on we apply the second step until
we’ve shown that every case is true
...
6
for n = 1, 2, 3,
...
Note that De Moivre’s formula deals with powers of certain
LECTURE 23
...
These are the same kind of
terms that we came across when we were dealing with trigonometric representation
of complex numbers
...
Example 1 Find the exact value for (1 + i)10
...
Doing this we will get the
following
...
In particular we can use De Moivre’s formula
to find equations for sine and cosine of multiples of an angle in terms of sine and
cosine of the angle
...
Solution Putting n = 2 into De Moivre’s formula we get the following
...
cos(2θ) = cos2 (θ) − sin2 (θ)
sin(2θ) = 2 sin(θ) cos(θ)
LECTURE 23
...
7
169
Finding roots
One of the most useful applications of De Moivre’s formula is finding nth roots
of complex numbers
...
If we write z in
trigonometric form, i
...
z = r(cos(θ) + i sin(θ)), then one of the nth roots of z is
given by the following
...
For example
there are two square roots and so we would expect that in general that there would
be a total of n, nth roots of z
...
Recall that in representing a number in trigonometric form that the angle θ is
not unique but that we can add any multiple of 360◦ (or 2π rads) to the angle and
get other representations for z
...
Specifically, if we have z = r(cos(θ) + i sin(θ)) then there are n nth roots of z
and they are given by the following
...
, n − 1
n
n
Note that if we put in k = n then this will correspond to the same root as when
k = 0
...
(The roots
will just keep repeating every time we go through a cycle of n values
...
Example 3 Find the cube roots of i
...
So in particular
by applying our formula for roots above (with n = 3) we will get the
following
...
√
3 1
k=0:
+ i
2√ 2
3 1
k=1:
−
+ i
2
2
k=2:
−i
LECTURE 23
...
23
...
Using De Moivre’s formula find equations for cos(3x) and sin(3x) in terms
of cos(x)’s and sin(x)’s
...
Find (1 + 3i)15
...
Find all the cube roots of −4 2−4 2i
...
4
...
Write your answers in trigonometric
form
...
Using induction show that 1 + 2 + · · · + n = n(n + 1)/2
...
6
...
2
2
Using mathematical induction prove this relationship is true for n = 1, 2,
...
Appendix A
Collection of equations
Over the course of these lectures we have encountered a large number of relationships
...
Reciprocal identities
csc(θ) =
1
,
sin(θ)
sec(θ) =
1
,
cos(θ)
cot(θ) =
1
,
tan(θ)
sin(θ) =
1
,
csc(θ)
cos(θ) =
1
,
sec(θ)
tan(θ) =
1
...
sin(θ)
Pythagorean identities
cos2 (θ) + sin2 (θ) = 1,
1 + tan2 (θ) = sec2 (θ),
cot2 (θ) + 1 = csc2 (θ)
...
Even/odd’er identities
cos(−θ) = cos(θ),
sin(−θ) = − sin(θ),
tan(−θ) = − tan(θ)
...
171
APPENDIX A
...
1 + tan(x) tan(y)
Double angle identities
sin(2x) = 2 sin(x) cos(x),
cos(2x) = cos2 (x) − sin2 (x) = 2 cos2 (x) − 1 = 1 − 2 sin2 (x)
...
2
Half-angle identities
r
r
x
x
1 + cos(x)
1 − cos(x)
=±
,
sin
=±
,
cos
2
2
2
2
x 1 − cos(x)
sin(x)
tan
=
=
...
2
cos(x) cos(y) =
APPENDIX A
...
2
2
The identity with no name
√
a sin(x) + b cos(x) = ( a2 + b2 ) sin(x + θ)
arccos √ 2a 2
if b ≥ 0
a +b
where θ =
360◦ − arccos √ 2a 2
if b < 0
a +b
g
a
b
b
a
c
Law of sines
b
c
a
=
=
sin(α)
sin(β)
sin(γ)
Law of cosines
a2 = b2 + c2 − 2bc cos(α)
b2 = a2 + c2 − 2ac cos(β)
c2 = a2 + b2 − 2ab cos(γ)
or
or
or
cos(α) = (b2 + c2 − a2 )/(2bc),
cos(β) = (a2 + c2 − b2 )/(2ac),
cos(γ) = (a2 + b2 − c2 )/(2ab)
...
2
APPENDIX A
...
x→0
x
lim
ha, bi · hc, di = ac + bd, ~u · ~v = ~v · ~u,
~u · (~v + w)
~ = ~u · ~v + ~u · w,
~
~u · ~v = k~ukk~v k cos(θ), ~u · ~u = k~uk2 ,
cos(θ) =
~u · ~v
...
◦
arctan(b/a) + 180
if a < 0
√
a2 + b 2
De Moivre’s formula
(cos(θ) + i sin(θ))n = cos(nθ) + i sin(nθ),
√
θ + k · 360◦
θ + k · 360◦
(1/n)
n
(r[cos(θ) + i sin(θ)])
= r cos
+ i sin
n
n
for k = 0, 1,
...
Title: Trigonometry - Brief And It's Formulas
Description: The trigonometric ratios are the ratios between the two sides of a right-angles triangle with respect to an angle and hence they are real numbers. The angle θ taken into consideration may be acute, obtuse or right angle. Various trigonometry formulas related to the basic trigonometric ratios which must be remembered include: 1) sin2 θ + cos2 θ = 1 2) sec2 θ -tan2 θ = 1 3) cosec2 θ - cot2 θ = 1 4) sin θ cosec θ = tan θ cot θ = sec θ cos θ = 1 5) sin2 θ + cos2 θ = 1, so each of them is numerically less than 1. 6) |sin θ| ≤ 1 and |cos θ| ≤ 1 7) -1
Description: The trigonometric ratios are the ratios between the two sides of a right-angles triangle with respect to an angle and hence they are real numbers. The angle θ taken into consideration may be acute, obtuse or right angle. Various trigonometry formulas related to the basic trigonometric ratios which must be remembered include: 1) sin2 θ + cos2 θ = 1 2) sec2 θ -tan2 θ = 1 3) cosec2 θ - cot2 θ = 1 4) sin θ cosec θ = tan θ cot θ = sec θ cos θ = 1 5) sin2 θ + cos2 θ = 1, so each of them is numerically less than 1. 6) |sin θ| ≤ 1 and |cos θ| ≤ 1 7) -1