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Section 5.2 - Graphs of the Sine and Cosine Functions In this section, we will graph the basic sine function and the basic cosine function and then graph other sine and cosine functions using transformations. Much of what we will do in graphing these problems will be the same as earlier graphing using transformations. Definition: A non-constant function f is said to be periodic if there is a number p > 0 such that
( ) ( )f x p f x+ = for all x in the domain of f. The smallest such number p is called the period of f. The graphs of periodic functions display patterns that repeat themselves at regular intervals. Definition: For a periodic function f with maximum value M and minimum value m.
The amplitude of the function is: .2
mM −
In other words the amplitude is half the height. Example 1: State the period and amplitude of the periodic function.
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Example 2: State the period and amplitude of the periodic function:
Note: For a periodic function f, the period of the graph is the length of the interval needed to draw one complete cycle of the graph. For a basic sine or cosine function, the period is .2π For a basic sine or cosine function, the maximum value is 1 and the minimum value is -1, so the amplitude
is .12
)1(1 =−−
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We’ll start with the graph of the basic sine function, )sin()( xxf = . The domain of this function is ( )∞∞− , and the range is [-1, 1]. We typically graph just one complete period of the graph, that is on the interval [ ]π2,0 . We’ll make a table of values: x 0
6
π
4
π
3
π
2
π
3
2π
4
3π
6
5π
π 6
7π
4
5π
3
4π
2
3π
3
5π
4
7π
6
11π
π2
sinx
0
2
1
2
2
2
3
1
2
3
2
2
2
1
0
2
1−
2
2−
2
3−
-1
2
3−
2
2−
2
1− 0
Then using these ordered pairs, we can sketch a graph of the function.
Next, we’ll draw in a smooth curve:
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Drawing all of these points is rather tedious. We’ll ask you to learn the shape of the graph and just graph five basic points, the x and y intercepts and the maximum and the minimum.
Period: 2π Amplitude: 1 x-intercepts: ,2π π y-intercept: (0,0)
Big picture: )sin()( xxf =
Period: 2π Amplitude: 1 x-intercepts: kπ y-intercept: (0,0)
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Now we’ll repeat the process for the basic cosine function )cos()( xxf = . The domain of this function is
( )∞∞− , and the range is [-1, 1]. Again, we typically graph just one complete period of the graph, that is on
the interval [ ]π2,0 . Here is the table of values for )cos()( xxf = : x 0
6
π
4
π
3
π
2
π
3
2π
4
3π
6
5π
π 6
7π
4
5π
3
4π
2
3π
3
5π
4
7π
6
11π
π2
cosx
1
2
3
2
2
2
1
0
2
1−
2
2−
2
3−
-1
2
3−
2
2−
2
1−
0
2
1
2
2
2
3
1
Now we’ll graph these ordered pairs.
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For the basic cosine graph, you’ll need to remember the basic shape and graph the x and y intercepts as well as the maximum and minimum points. Period: 2π
Amplitude: 1
x-intercepts: 3
,2 2
π π
y-intercept: (0,1) Big picture: )cos()( xxf = Period: 2π
Amplitude: 1
x-intercepts: 2
kπ ( k is an odd integer)
y-intercept: (0,1)
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Now we’ll turn our attention to transformations of the basic sine and cosine functions. These functions will be of the form .)cos()(or)sin()( DCBxAxgDCBxAxf +−=+−= We can stretch or shrink sine and cosine functions, both vertically and horizontally. We can reflect them about the x axis, the y axis or both axes, and we can translate the graphs either vertically, horizontally or both. Next we’ll see how the values for A, B, C and D affect the graph of the sine or cosine function.
Graphing ( ) sin( ) or ( ) cos( )f x A Bx C D g x A Bx C D= − + = − +
• The amplitude of the graph of is A .
• The period of the function is: .2B
π
• If A > 1, this will stretch the graph vertically.
0 < A < 1, this will shrink the graph vertically If A < 0, the graph will be a reflection about the x axis.
• If B >1 , this will shrink the graph horizontally by a factor of 1/B. If 0<B<1, this will stretch the graph horizontally by a factor of 1/B.
• Vertical Shift: Shift the original graph D units UP if D > 0, D units DOWN if D < 0.
• Phase shift: The function will be shifted B
C units to the right if 0>
B
Cand to the left if 0<
B
C. The
number B
Cis called the phase shift.
Note: Horizontal Shift: If the function is of the form )sin()( Cxxf −= or )cos()( Cxxf −= , then shift the original graph C units to the RIGHT if C > 0 and C units to the LEFT if C < 0.
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Example 3: Write down the transformations needed to graph:
( ) sin(4 )f x x= Period:
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Amplitude:
Transformations:
1( ) cos
2f x x
=
Period:
Amplitude:
Transformations:
( )( ) 2cosf x x π= − Period:
Amplitude:
Transformations:
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( )( ) 5sin 4f x x π= − + Period:
Amplitude:
Transformations:
( )( ) sin 2 1f x x π= − + Period:
Amplitude:
Transformations:
( ) 5sin 12 8
xf x
π π = − −
Period:
Amplitude
Transformations:
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It can be helpful to identify the starting and ending points for one period of the graph of a function that has a phase shift. To do this, solve the equations .2 and 0 π=−=− CBxCBx
( )( ) 5cos 2f x x π= − ; starting point: 2 02
x xππ− = → =
ending point: 3
2 2 2 32
x x xππ π π− = → = → =
You will need to identify the transformations required to change a basic sine or cosine function to the desired one. You must know the five key points on a basic sine function and the five key points on a basic cosine function. Using the information about the amplitude, reflections, vertical and horizontal stretching or shrinking and vertical and horizontal translations, you will be able to correctly plot the translated key points and sketch the desired function. Example 4: Sketch over one period: )sin(4)( xxf =
13
Example 5: Sketch over one period: ( ) cos2
xf x
= −
14
Example 6: Sketch over one period: ( ) 4cos(2 ) 1f x xπ= −
15
Example 7: Sketch over one period: ( ) sin 2 12
f x xπ = + +
16
Example 8: Sketch over one period: 1
( ) 2cos4 8
f x xπ = − −
17
Exercise: Sketch over one period: ( ) 2sin 12 8
xf x
π π = − −
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Example 9: Consider the graph: Write an equation of the form DCBxAxf +−= )sin()( and an equation of the form DCBxAxf +−= )cos()( which could be used to represent the graph. Note: these answers are not unique!
Example 10: Consider the graph: Write an equation of the form DCBxAxf +−= )sin()( and an equation of the form DCBxAxf +−= )cos()( which could be used to represent the graph. Note: these answers are not unique!
19
Exercise: Consider the graph: Write an equation of the form DCBxAxf +−= )sin()( and an equation of the form DCBxAxf +−= )cos()( which could be used to represent the graph. Note: these answers are not unique!
1
Section 5.3a - Graphs of Secant and Cosecant Functions
Using the identity ,)sin(
1)csc(
xx = you can conclude that the graph of g will have a vertical
asymptote whenever .0)sin( =x This means that the graph of g will have vertical asymptotes at …,2,,0 ππ ±±=x . The easiest way to draw a graph of )csc()( xxg = is to draw the graph of
)sin()( xxf = , sketch asymptotes at each of the zeros of )sin()( xxf = , then sketch in the cosecant graph.
1( ) csc( )
sin( )g x x
x= = ; if sin( ) 0x = , then g(x) has a vertical asymptote.
Here’s the graph of )sin()( xxf = on the interval
−2
5,
2
5 ππ.
Next, we’ll include the asymptotes for the cosecant graph at each point where .0)sin( =x
2
Now we’ll include the graph of the cosecant function.
Period: 2π Vertical Asymptote: x = kπ , k is an integer x-intercepts: None y-intercept: None Domain: x ≠ kπ , k is an integer Range: (−∞,−1]∪[1,∞) Typically, you’ll just graph over one period ( )π2,0 . To graph csc( )y A Bx C D= − + , first graph, THE HELPER GRAPH: sin( )y A Bx C D= − + .
3
You’ll also be able to take advantage of what you know about the graph of )cos()( xxf = to help
you graph )sec()( xxg = . Using the identity ,)cos(
1)sec(
xx = you can conclude that the graph of
g will have a vertical asymptote whenever .0)cos( =x
This means that the graph of g will have vertical asymptotes at …,2
3,
2
ππ ±±=x . The easiest
way to draw a graph of )sec()( xxg = is to draw the graph of )cos()( xxf = , sketch asymptotes at each of the zeros of )cos()( xxf = , then sketch in the secant graph.
1( ) sec( )
cos( )g x x
x= = ; if cos( ) 0x = , then g(x) has a vertical asymptote.
Here’s the graph of )cos()( xxf = on the interval .2
5,
2
5
− ππ
Next, we’ll include the asymptotes for the secant graph.
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Now we’ll include the graph of the secant function.
Period: 2π Vertical Asymptote:2kπ
/ 2x kπ= is an odd integer x-intercepts: None y-intercept: (0, 1) Domain: / 2x kπ≠ , k is an odd integer Range: (−∞,−1]∪[1,∞) Typically, you’ll just graph over one period ( )π2,0 . To graph sec( )y A Bx C D= − + , first graph, THE HELPER GRAPH: cos( )y A Bx C D= − + .
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Example 1: Sketch ( ) 4sec2
xf x
=
6
Example 2: Sketch ( ) 2csc2 2
xf x
π π = − −
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Exercise: Find the vertical asymptotes of:
a) ( ) 2sec2x
f x ππππ = −= −= −= −
b) ( ) 2csc4
f x xππππ = −= −= −= −
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Section 5.3b - Graphs of Tangent and Cotangent Functions
Tangent function: sin( )
( ) tan( )cos( )
xf x x
x= = ;
Vertical asymptotes: when 0)cos( =x , that is …,2
5,
2
3,
2
πππ ±±±=x
Domain: 3 5
, , ,2 2 2
xπ π π≠ ± ± ± … Range: ( ),−∞ ∞
x-intercepts: when 0)sin( =x , that is …,2,,0 ππ ±±=x Period: π
2
Often you will need to graph the function over just one period. In this case, you’ll use the interval
−2
,2
ππ. Here’s the graph of )tan()( xxf = over this interval, with pertinent points marked.
To graph DCBxAxf +−= )tan()( ;
• The period is: B
π
• Find two consecutive asymptotes by solving: 2
Bx Cπ− = and
2Bx C
π− = − .
• Find an x-intercept by taking the average of the consecutive asymptotes.
• Find the x coordinates of the points halfway between the asymptotes and and the x-intercept.
Evaluate the function at these values to find two more points on the graph of the function.
Note: If B > 1, it’s a horizontal shrink. If 0 < B < 1, it’s a horizontal stretch.
3
Example 1: Sketch ( ) 2 tan4
xf x
=
4
Example 2: Sketch ( ) 2 tan4
f x xππ = −
5
Cotangent Function: cos( )
( ) cot( )sin( )
xf x x
x= = ;
Vertical asymptotes: when 0)sin( =x , that is …,2,,0 ππ ±±=x Domain: 0, , 2 ,x π π≠ ± ± … Range: ( ),−∞ ∞
x-intercepts: when 0)cos( =x , that is …,2
5,
2
3,
2
πππ ±±±=x
Period: π
6
Often you will need to graph the function over just one period. In this case, you’ll use the interval ( )π,0 . Here’s the graph of )cot()( xxf = over this interval.
You can take the graph of either of these basic functions and draw the graph of a more complicated function by making adjustments to the key elements of the basic function. The key elements will be the location(s) of the asymptote(s), x intercepts, and the translations of the
points at
−
−
1,4
3or 1,
4
-either and1,
4
πππ.
To graph DCBxAxg +−= )cot()( ;
• The period is: B
π
• Find two consecutive asymptotes by solving: 0Bx C− = and Bx C π− = . • Find an x-intercept by taking the average of the consecutive asymptotes.
• Find the x coordinates of the points halfway between the asymptotes and and the x-intercept.
Evaluate the function at these values to find two more points on the graph of the function.
Note: If B > 1, it’s a horizontal shrink. If 0 < B < 1, it’s a horizontal stretch
8
Example 4: Sketch ( ) 5cot(2 )f x x=
9
Example 5: Give an equation of the form f (x) = Atan(Bx - C)+ D and f (x) = Acot(Bx - C)+ D that could represent the following graph.
Exercise: Give an equation of the form f (x) = Atan(Bx - C)+ D and f (x) = Acot(Bx - C)+ D that could represent the following graph.
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