Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
Trigonometry
Chapter 4 Lecture Notes
Section 4.1
I.
Graphs of the Sine and Cosine Functions
Periodic Functions
A periodic function is one whose function values repeat in a regular pattern. They are used
to describe cyclical phenomena. All six trigonometric functions are periodic.
The smallest increment containing one full repetition of the pattern is called the period of the
function.
II.
The Graph of the “Plain Vanilla” Sine Function
The graph of the “plain vanilla” sine function, y = sin x, has the following characteristics:
a.
Domain: (-∞, ∞)
b.
Amplitude: 1
Range: [-1, 1]
Note: The amplitude is the height above or the depth below the graph’s horizontal axis.
It is half the distance between the maximum and minimum y-values.
c.
d.
π
2
π
2π
5 key points: (0, 0)  , 1  , 0 
2   2 
Period: 2π;
x-increment:
Note: The x-increment is usually
1
of the period.
4
 3π
  4π 
 , - 1  , 0  .
2

  2 
Note that its y-values start in the “neutral position” with y1 = 0.
e.
Sine is an odd function with origin symmetry. For odd functions, f(–x) = –f(x), so for
sine, sin (–x) = –sin x. If the point (a, b) is on the graph of an odd function, the point (–a, –b)
is also on the graph.
1
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
III.
The Graph of the “Plain Vanilla” Cosine Function
The graph of the “plain vanilla” cosine function, y = cos x, has the following characteristics:
a.
Domain: (-∞, ∞)
b.
Amplitude: 1
c.
Period: 2π;
d.
Range: [-1, 1]
π
2
π
2π
3π
4π
5 key points: (0, 1)  , 0   , - 1  , 0   , 1 .
2
2
2

 
 
  2 
x-increment:
Note that its y-values start in the “up position” with y1 = 1.
e.
Cosine is an even function with y-axis symmetry. For even functions, f(–x) = f(x), so for
cosine, cos (–x) = cos (x). Thus, if the point (a, b) is on the graph of an even function, the
point (–a, b) is also on the graph.
IV. Transformations of the “Plain Vanilla” Sine and Cosine Functions
A.
Amplitude and x-axis reflection
In the functions y = a sin x and y = a cos x, the a-value controls two things – the amplitude of
the graph and whether there has been an x-axis reflection.
The amplitude of the graph is a .
If a < 0, the graph has been reflected over the x-axis.
B.
Period and x-increment
In the functions y = sin bx and y = cos bx, the b-value controls the period, which in turn
controls the x-increment.
The period is
2π
b
and the x-increment is
1
 period .
4
2
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
Example 1
Find the amplitude, period, and x-increment for each of the given functions and
identify whether there has been an x-axis reflection.
1
b. y = –5 sin πx
4
1
1
 amp = ; there is no x-axis reflection.
a. a =
4
4
2π
1 2π π
b = 3  period =
; x-increment =  
4 3
6
3
a. y  cos 3x
b. a = –5  amp = 5; there is an x-axis reflection
b = π  period =
Example 2
Graph two full periods of y = 3 sin 2x. Give a, b, amplitude, whether there is an
x-axis reflection, period, x-increment, and 5 key points.(#24)
a=3
amp = 3
b=2
period =
(0, 0)
Example 3
2π
1
1
 2 ; x-increment =  2 
π
4
2
π 
 , 3
4 
no x-axis reflection
2π
π
2
 3π

 , - 3
 4

 2π 
 , 0
 4 
x-increment =
π
4
 4π 
 , 0
 4 
π
4
Graph two full periods of y  2 cos . Give a, b, amplitude, whether there is an
x-axis reflection, period, x-increment, and 5 key points.(#24)
a = –2
amp = 2
There is an x-axis reflection
π
b=
4
2π
4
period =
 2π   8
π
π
4
x-increment =
(0, –2)
(2, 0)
(4, 2)
(6, 0)
8
2
4
(8, –2)
3
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
Section 4.2
I.
Translations of the Graphs of the Sine and Cosine Functions
Phase Shifts (a.k.a. Horizontal Translations)
To find the phase shift of a function in the form y = a sin (bx – c) or y = a cos (bx – c),
set the argument of the function (what is inside parentheses, after the function name)
c
b
equal to zero and solve for x: bx  c  0  bx  c  x  .
II.
If
c
b
is positive (i.e. you see subtraction in the parentheses), the phase shift is
If
c
b
is negative (i.e. you see addition in the parentheses), the phase shift is
c
b
c
b
to the right.
to the left.
Vertical Shifts (a.k.a. Vertical Translations)
In the functions y = a sin (bx – c) + d and y = a cos (bx – c) + d, the d controls the vertical shift.
If d is positive, vertical the shift is d up. If d is negative, the vertical shift is d down.
Example 1
Find the amplitude, period, x-increment, phase shift, and vertical shift for each of the
given functions and identify whether there has been an x-axis reflection.
a. y =
1
cos (2x – 3π) – 1 (#22)
2
3
π
b. y = 2 – 3sin  x  
4
8
1
1
 amp = ; there is no x-axis reflection.
2
2
1
π
2π
 π ; x-increment =  π 
b = 2  period =
4
4
2
3π
3π
 phase shift:
c = 3π  2x – 3π = 0  2x = 3π  x =
right
2
2
a. a =
d = –1  vertical shift: 1 down
b. a = –3  amp = 3; there is an x-axis reflection
3
1 8π 2π
2π
4 8π
 period =
; x-increment =  
 2π  
3
4 3
3
4
3 3
4
π
3
π
3
π
π 4
π
π

x
c=
= 0  x    x       phase shift: left
8
4
8
4
8
8 3
6
6
b=
d = 2  vertical shift: 2 up
III.
How to Graph y = a sin (bx – c) + d and y = a cos (bx – c) + d
1.
Identify a, b, c, and d.
2a.
Find the amplitude: amp = |a|.
b.
3a.
Determine if there has been an x-axis reflection: if a is negative, there has been an x-axis
reflection.
Find the period: period = 2π .
b
4
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
b.
4.
Find the x-increment: x-increment =
1
 period .
4
Find the phase shift by setting the argument, bx – c, equal to zero and solving for x:
c
c
right (if we see bx – c and is positive) or
b
b
c
c
P.S. =
left (if we see bx + c and is negative).
b
b
P.S. =
5.
Find the vertical shift: |d| up if d is positive or |d| down if d is negative.
6.
Identify the five key points:
c
for sine:  , d 
b

(x1 + x-inc, d  a )
c
for cosine:  , d  a 
b
7.

(x2 + x-inc, d) (x3 + x-inc, d – a)
(x1 + x-inc, d) (x2 + x-inc, d – a)
(x4 + x-inc, d)
(x3 + x-inc, d) (x4 + x-inc, d + a)
Plot the five key points and connect them with a smooth, continuous curve, then extend the graph
to two full periods.
Example 2: y = –3 sin (2x – π) + 1
1.
a = –3
b=2
c=π
2.
amp = |–3| = 3
3.
period =
4.
phase shift =
5.
vertical shift = 1 up
6.
5 key points: 
d=1
There is an x-axis reflection.
2π
π
2
x-increment =
π
2π
or
right
2
4
 2π 
, 1
 4

 3π

 , - 2
 4

1
π
π 
4
4
2x  π  0  2x  π  x 
 4π 
 , 1
 4

 5π

 , 4
 4

π 2π

2
4
 6π 
 , 1
 4

7.
5
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
Example 3: y =
1
3
1
3
cos (4x + π) – 2
c = –π
1.
a=
b=4
2.
amp =
3.
period =
4.
phase shift =
5.
vertical shift = 2 or
6.
5 key points:  
1
3
=
1
3
d = –2 = –
6
3
There is no x-axis reflection.
2π π

4
2
x-increment =
π
2π
or
left
8
4
6
3
1 π π
 
4 2 8
4x  π  0  4x   π  x  
π
2π

4
8
down
5
 2π
,  
3
 8
6
 π
 ,  
3
 8
7

 0,  
3

6
π
 ,  
3
8
5
 2π
,  

3
 8
7.
6
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
Section 4.3
Graphs of the Other Circular Functions
Note: The concept of amplitude does not apply to cosecant, secant, tangent, or cotangent. We can only
discuss the amplitude of a sine or cosine function.
I.
The Basic Cosecant Function
π
The graph of y = csc x has a period of 2π and an x- increment of 2 . It is an odd function with
1
origin symmetry. Because csc x 
, the cosecant function is undefined whenever sin x = 0.
sin x
Thus the domain of csc x is all real numbers except integer multiples of π. At x = nπ, where n is
an integer, the graph of y = sin x crosses the horizontal axis and the graph of y = csc x has a
vertical asymptote. The range of csc x is (-∞, 1] U [1, ∞).
The general procedure for graphing y = a csc (bx – c) + d is to graph y = a sin (bx – c) + d with a
dotted line and then to use that graph as a guide for graphing the cosecant. The graph of
y = csc x has vertical asymptotes at x = 0, x = π, and x = 2π. It has a minimum (low point) at
π 
 3π

 , 1 and a maximum (high point) at  , - 1 .
2 
 2

II.
The Basic Secant Function
The graph of y = sec x has a period of 2π and an x- increment of
π
. It is an even function with
2
1
, the secant function is undefined whenever cos x = 0.
cos x
π
π
Thus the domain of sec x is all real numbers except odd multiples of . At x   n  π , where n
2
2
y-axis symmetry. Because sec x 
is an integer, the graph of y = cos x crosses the horizontal axis and the graph of y = sec x has a
vertical asymptote. The range of sec x is (-∞, 1] U [1, ∞).
The general procedure for graphing y = a sec (bx – c) + d is to graph y = a cos (bx – c) + d with a
dotted line and then to use that graph as a guide for graphing the secant. The graph of y = sec x
π
3π
and x  . It has minima (low points) at (0, 1) and (2π, 1).
2
2
It has a maximum (high point) at π, - 1 .
has vertical asymptotes at x 
7
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
III.
How to Graph y = a csc (bx – c) + d and y = a sec (bx – c) + d
1.
Write the guide function. For y  a csc (bx  c)  d the guide function is y  a sin (bx  c)  d and
for y  a sec (bx  c)  d the guide function is y  a cos (bx  c)  d .
2.
Identify a, b, c, and d.
3a.
Find the amplitude of the guide function: amp = |a|.
Note: Cosecant and secant do not have amplitude.
b.
4a.
b.
5.
Determine if there has been an x-axis reflection: if a is negative, there has been an x-axis
reflection.
Find the period: period =
2π
b
.
Find the x-increment: x-increment =
1
 period .
4
Find the phase shift by setting the argument, bx – c, equal to zero and solving for x:
c
c
right (if we see bx – c and is positive) or
b
b
c
c
P.S. =
left (if we see bx + c and is negative).
b
b
P.S. =
6.
Find the vertical shift: |d| up if d is positive or |d| down if d is negative.
7.
Identify the five key points of the guide function:
c
for sine:  , d 
b

(x1 + x-inc, d+a)
c
for cosine:  , d  a 
b

(x2 + x-inc, d) (x3 + x-inc, d – a)
(x1 + x-inc, d) (x2 + x-inc, d – a)
(x4 + x-inc, d)
(x3 + x-inc, d) (x4 + x-inc, d + a)
8.
Plot the five key points of the guide function and connect them with a smooth, continuous,
dotted curve, then extend the graph to two full periods.
9.
Draw the given function using these guidelines:
Anywhere the guide function crosses its horizontal axis (i.e., when y = d), the given function has
a vertical asymptote. Anywhere the guide function has a minimum, the given function has a
maximum and we draw a parabola shaped curve opening down. Anywhere the guide function
has a maximum, the given function has a minimum, and we draw a parabola shaped curve
opening up.
8
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
Example 1: y = 2 csc (x – 3π) – 1
1.
Guide function: y = 2 sin (x – 3π) – 1
2.
a=2
3.
amp of sine = |2| = 2
4.
period =
5.
phase shift = 3π or
6.
vertical shift = 1 down
7.
6π
5 key points of the sine:  ,  1
b=1
6π
2
c = 3π =
2π
 2π
1
d = –1
There is no x-axis reflection.
x-increment =
1
2π π
 2π 

4
4
2
6π
right
2
x  3π  0  x  3π
 2
 7π 
 , 1
 2


8.
 8π

 ,  1
 2

 9π

 ,  3
 2

 10π

, 1

 2

9.
Example 2: y = –
π
1
sec  x   + 3
4
2

π
1
cos  x   + 3
4
2

1.
Guide function: y = –
2.
a=–
3.
amp of cosine = 
4.
period =
5.
phase shift =
6.
vertical shift = 3 or
7.
π 11
5 key points of the cosine:   , 
1
4
b=1
c=–
2π
 2π
1
π
2
1
1
=
4
4
d=3=
There is an x-axis reflection.
x-increment =
π
left
2
12
4
x
1
2π π
 2π 

4
4
2
π
0
2

x
π
2
12
up
4
 2
4
 12 
 0,

4

 π 13 
 ,

2 4 
 2π 12 
 ,

 2 4
 3π 11 
 , 
 2 4
9
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
8.
IV.
9.
The Basic Tangent Function
The graph of y = tan x has a period of π and an x- increment of
π
. It is an odd function with
4
sin x
, the tangent function is undefined whenever cos x = 0.
cos x
π
π
Thus the domain of tan x is all real numbers except odd multiples of . At x   n  π , where n
2
2
origin symmetry. Because tan x 
is an integer, the graph of y = tan x has a vertical asymptote. The range of tan x is (-∞, ∞).
Notice that the graph of y = tan x is increasing along its entire length. Its left “hand” points
down and its right hand points up. The 5 key features of the graph of y = tan x are:
left asymptote: x  
V.
π
2
π
π
3 key points:   ,  1 (0, 0)  , 1

4

4

right asymptote: x 
π
2
The Basic Cotangent Function
The graph of y = cot x has a period of π and an x- increment of
origin symmetry. Because cot x 
cos x
sin x
π
. It is an odd function with
4
, the cotangent function is undefined whenever sin x = 0.
Thus the domain of cot x is all real numbers except integer multiples of π. At x  n  π , where n is
an integer, the graph of y = cot x has a vertical asymptote. The range of cot x is (-∞,∞).
Notice that the graph of y = cot x is decreasing along its entire length. Its left “hand” points up
and its right hand points down. The 5 key features of the graph of y = cot x are:
10
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
left asymptote: x = 0
VI.
π
π
3π
3 key points:  , 1  , 0   , - 1
4
 2
  4
right asymptote: x  π

How to Graph y = a tan (bx – c) + d and y = a cot (bx – c) + d
1.
Identify a, b, c, and d.
2.
Determine if there has been an x-axis reflection: if a is negative, there has been an x-axis
reflection.
3a.
Find the period: period =
b.
4.
π
b
.
Find the x-increment: x-increment =
1
 period .
4
Find the phase shift by setting the argument, bx – c, equal to zero and solving for x:
P.S. =
c
b
right (if we see bx – c and
P.S. =
c
b
left (if we see bx + c and
c
b
c
b
is positive) or
is negative).
5.
Find the vertical shift: |d| up if d is positive or |d| down if d is negative.
6.
Identify the left asymptote, the three key points, and the right asymptote.
c
b
cotangent: To find the left asymptote, let x = phase shift, x  .
LA: x 
c
b
(x of LA + x-increment, d + a) (x1 + x-inc, d) (x2 + x-inc, d – a) RA: x = x3 + x-inc
Note: The left asymptote of a tangent function is NOT x = phase shift.
tangent: To find the left asymptote, set the argument, bx – c, equal to 
c
b
LA: x  
7.
π
2b
π
and solve for x.
2
(x of LA + x-inc, d – a) (x1 + x-inc, d) (x2 + x-inc, d + a) RA: x = x3 + x-inc
Plot the asymptotes and the three key points. Connect the points with a smooth, continuous curve.
Extend the graph to two full periods.
11
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
1
Example 3: y = 2 tan  x  π  – 3
4
b=
1
4

c=π
d = –3
1.
a=2
2.
There is no x-axis reflection.
3.
period =
π
1
 4π
x-increment =
4
4.
phase shift = 4π right
5.
vertical shift = 3 down
6.
LA:
1
 4π  π
4
1
xπ0
4

1
π
xπ
 x  4π  2π  x  2 π
4
2
x  4π  0 
(3π, –5)
x  4π
(4π, –3)
(5π, –1)
RA: x = 6π
7.
Example 4: y = –3 cot (2x + π) +1
1.
a = –3
2.
There is an x-axis reflection.
3.
period =
4.
phase shift =
5.
vertical shift = 1 up
6.
LA: x  
b=2
π
2
c = –π
d=1
x-increment =
π
2
π
left
2

4π
8
2x  π  0  2x  π  x  
 3π

,  2

 8

 2π 
, 1

 8

 π 
  , 4
 8 
π
4π

2
8
RA: x = 0
7.
12
Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
Section 4.4
I.
Harmonic Motion
Definition
Simple harmonic motion is a periodic, oscillating movement which can be described by a
sinusoidal function. Examples of physical phenomena which can be described as simple
harmonic motion include radio and television waves; light waves; sound waves; water waves;
the swinging of a pendulum; the vibrations of a tuning fork; and the bobbing of a weight attached
to a coiled spring.
II.
The Movement of a Mass on a Spring
Consider a mass on a spring. The system is said to be in equilibrium when the mass is at rest.
The point of rest is called the origin of the system. We consider the distance above equilibrium
as positive and the distance below equilibrium as negative.
If the mass is lifted a distance a and released, it will oscillate up and down in periodic motion.
If there is no friction, the motion repeats itself in a certain period of time. The distance a is
called the displacement from the origin. The number of times the mass oscillates in 1 unit of
time is called the frequency f of the motion, and the time it takes for the mass to complete one
1
oscillation is called its period. The frequency and period are related by the formulas: f  period
and
period 
1
f
. The maximum displacement from equilibrium is called the amplitude of the
motion.
III.
Equations Modeling Simple Harmonic Motion
If the displacement from the origin is at a maximum at time t = 0, simple harmonic motion can
be modeled by y = a cos ωt.
If the displacement from the origin is zero at time t = 0, simple harmonic motion can be modeled
by y = a sin ωt.
For either model, | a | = amplitude (maximum displacement); p = period =
f = frequency =
ω
2π
; ω = 2πf or ω 
2π
ω
;
2π
; t is the time, and y is the displacement at time t.
p
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Ch 4 Notes F07 O’Brien
LHS Trig 8th ed
Example 1
An object is attached to a coiled spring. It is pulled down a distance of 6 units from its
equilibrium position, and then released. The time for one complete oscillation is 4 sec.
(#10)
a. Give an equation that models the position of the object at time t.
a = -6
p=4
ω=
2π π

4
2
y  6 cos
π
t
2
b. Determine the position at t = 1.25 sec.
y  6 cos
π
(1.25)  2.30 units
2
c. Find the frequency.
f
1 1

p 4
The formula for the up and down motion of a weight on a spring is given by y  a sin
k
t where
m
k is the spring constant and m is the mass of the weight.
Example 2
A spring with a spring constant of k = 3 has a 27 unit mass suspended from it.
a. If the spring is stretched 1.5 ft and released, what are the amplitude, period
and frequency of the resulting oscillatory motion?
a = 1.5; b =
3
1

27 3
; amp = 1.5; period =
b. What is the equation of the motion?
Example 3
2π
1
 6π ;
frequency =
3
1
6π
3
1
y  sin t
2
3
The height of a weight attached to a spring is y = –4 cos 8πt inches after
t seconds. (Ch 4 Test #9)
a. Find the maximum height that the weight rises above the equilibrium position
of y = 0.
max height = amplitude = 4 inches
b. When does the weight first reach it maximum height, if t ≥ 0?
4 = –4 cos 8πt
8πt = π


t
–1 = cos 8πt
π 1

8π 8

cos θ = –1 when θ = π

second
c. What are the frequency and period?
f
b
8π

 4 cycles
2π 2π
per second; period =
1
second
4
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