Section 4.5
Sine and Cosine Parent Graphs
Sine Parent Function Graph
y = sin x is the parent function for the
sine function graph.
Its domain is (−∞, ∞) in both degrees and
radians.
Its range is [-1, 1] using the unit circle.
Find the following ordered pairs for the
sine function.
y = sin x
(0, 0),   , 1  , (, 0 ),  3 ,  1  , (2, 0)
2 
 2





 
 , (–, 0 ),  3  , (–2, 0 )
  2 ,  1
  2 , 1




These are called the key points of the
parent function graph. Now graph them on
the given graph.
y
y = sin x

x













Cosine Parent Function Graph
y = cos x is the parent function for the
cosine function graph.
Its domain is (−∞, ∞) in both degrees and
radians.
Its range is [-1, 1] using the unit circle.
Find the following ordered pairs for the
cosine function.
y = cos x
 3



(0, 1 ),  , 0  , (, -1 ),  , 0 , (2, 1 )
 2

2

 
 2, 0


 3

, (–, -1 ),   , 0  , (–2, 1 )

 2

Then graph them on the given graph.
y
y = cos x

x













A periodic function is a function that
repeats itself within the same interval. A
function f(x) is periodic if there exists a
smallest positive p such that
f(x + p) = f(x) whenever both f(x + p) and
f(x) are defined. The number p is said to
be the period of the function, and one
cycle of the graph is completed in each
period.
Use the graph of y = sin x to answer the
following questions.
1.
Is y = sin x periodic? yes
2.
What is the period?
3.
Is the function even or odd?
4.
Write the domain of the function in
interval notation. (-∞, ∞)
5.
Write the range of the function in
interval notation. [-1, 1]
2
odd
Use the graph of y = cos x to answer the
following questions.
1.
Is y = cos x periodic? yes
2.
What is the period?
3.
Is the function even or odd?
4.
Write the domain of the function in
interval notation. (-∞, ∞)
5.
Write the range of the function in
interval notation. [-1, 1]
2
even
Using Transformations on the
Sine and Cosine Graphs
The equations for sine and cosine functions
with transformations are
y = a sin b(x – c) + d
y = a cos b(x - c) + d
where,
a = the amplitude or height of the graph
b = horizontal stretch or shrink
c = phase shift(horizontal shift)
d = vertical shift
The period for a sine or cosine function is
2
.
b
When graphing a sine or cosine function
there will always be 5 key points in one
cycle.
The distance between each key point on
the x-axis is:
period
4
Examples
Find the amplitude, period, phase shift,
and vertical shift. Graph two periods
(two cycles).
1.
y = 5 sin 2x + 3
amplitude = 5
y = 5 sin 2x + 3
2π
π
period =
2
π
The key points are
units apart.
4
phase shift = 0
vertical shift = 3 up
The vertical shift is called the midline of
the new graph.
To find the beginning and ending points for
the two cycles:
1.
Subtract the period from the phase
shift to find the beginning point.
2. Add the period to the phase shift to
find the ending point.
beginning point: 0  π   π
ending point: 0  π  π
y = 5 sin 2x + 3

y







x

















π
2. y  2 sin3  x  
6

Remember there is a reflection across
the x-axis.
amplitude = 2
2π
period 
3
π
The key points are
units apart.
6
π
phase shift  right
6
vertical shift = 0
π 2π
π

beginning point: 
6 3
2
π 2π 5π

ending point: 
6 3
6
y




2


3


6

6



3

2
2
3
5
6
x

1
π
3. y  cos  x    1
2
4

1
amplitude 
2
period = 2π
π
The key points are
units apart.
2
π
phase shift  to the right
4
vertical shift= 1 down
π
7π
 2π  
beginning point :
4
4
π
9π
 2π 
ending point :
4
4
y

7

4
5

4
3

4


4




4
3
4
5
4
7
4
x
9
4
4.
y = 2 cos(3x + ) − 4

π
rewrite as y  2cos3  x    4
3

amplitude = 2
2π
period 
3
π
The key points are
units apart.
6
π
phase shift  to the left
3
vertical shift = 4 units down
π 2π
 π
beginning point :  
3
3
π 2π π

ending point :  
3
3
3
y

5

6
2

3


2


3

6


6







3
x
5.
y = 1 + 3cos (πx + 4π)
Rewrite: y = 3cos π (x + 4) + 1
amplitude = 3
2π
period 
2
π
1
The key points are
units apart.
2
phase shift = 4 to the left
vertical shift = 1 up
beginning point : 4  2  6
ending point : 4  2  2
y
4
3
2
6 5.5 5
4.5 4 3.5 3 2.5 2
x
1
2
Sinusoidal Application Problems
Example 1
Suppose that the waterwheel in the
figure on the next slide rotates at 6
revolutions per minute (rpm). Two
seconds after you start a stopwatch,
point P on the rim of the wheel is at its
greatest height, d = 13 ft., above the
surface of the water. The center of the
waterwheel is 6 ft. above the surface.
A.
1.
Sketch the graph of “d” as a function
of “t”, in seconds, since you started
the stopwatch.
On your graph sketch d = 6 ft. as your
midline. Now find the highest and
lowest point on the graph.
highest point = 6 + 7 = 13 feet
lowest point = 6 − 7 = −1 foot
The high point occurs at 2 seconds after
the stopwatch is started.
So sketch the 1st point at (2, 13).
(2, 13)
2.
Now find the period for your graph.
Since the waterwheel rotates at 6
rpms, then every 10 seconds the
waterwheel does 1 revolution. So
p = 10 seconds.
3.
Sketch a high point 10 seconds from
the 1st high point.
(12, 13)
(2, 13)
(12, 13)
4.
Mark a low point halfway between
these two high points.
(7, −1)
5.
Sketch the graph through these
points.
(12, 13)
(2, 13)
(7, −1)
B.
1.
Assuming that “d” is a sinusoidal
function of “t”, write a particular
equation.
Write the general equation. Use “d”
and “t” for the variables.
d = d + a cos b(t − c)
2.
Find a, b, c, and d.
d=6
3.
a=7
c=2
Write the equation.

d  6  7cos  t  2 
5
2 
b

10 5
C.
How high above or below the water’s
surface will P be at time t = 17.5 sec?
At that time will the waterwheel be
going up or down?

d  6  7 cos 17.5  2 
5
d  0.66 feet
Graph this equation in your calculator
to decide if the wheel is going up or
down.
The waterwheel is going up.
D.
At what time t was point P first
emerging from the water?

0  6  7 cos  t  2 
5
6 
1 
cos      t  2 
 7 5
 6
cos     2.60 or 3.68
 7
1
since cosine is negative in the 2nd and
3rd quadrants.
(12, 13)
(2, 13)
(7, −1)
We want the 3.68 because the 1st one
would give us the answer for the 1st time
d = 0 which is when the waterwheel is
going into the water and we need the value
for “t” when d = 0 when the waterwheel is
coming out of the water.

3.68   t  2 
5
t  7.9 sec.
Note that it is usually easier to use the
cosine function for these types of
problems because cosine starts at the
highest point.