Period of Trigonometric Functions
In previous lessons we have learned how to translate any primary trigonometric
function horizontally or vertically, and how to Stretch Vertically (change Amplitude).
In this unit we are going to learn how to Stretch Horizontally. In most “Real World”
applications that involve trigonometry, what we are doing is determining the
trigonometric function’s Period.
In order to master the techniques explained here it is very important that you
undertake plenty of practice exercises so that they become second nature.
After reading this unit, you should be able to complete the following:
 Determine the period for Sine, Cosine, and Tangent
 Sketch the function on blank grid
 Sketch the function on fixed grid
Recall:
The three basic trigonometric functions have periods as demonstrated below:
 Sine function -> period is 2 radians or 360
.
 Cosine function -> period is 2 radians or 360
.
 Tangent function -> period is  radians or 180.
The basic graphs of these 3 trigonometric functions are:
The length of one complete cycle of a trigonometric function is called the Period.
Typically we use x=0 as the starting point for the graph. When the length of the period
is not the default, the functions will be written in the format similar to:
f 
t sin 
kt ,
g 
x cos 
kx , or h 
 tan 
k . Where the constant k aids us in determining the
period for the function.
The Period (wavelength) of sin(kt) and cos(kt)
If k>0, then the graph of
f 
t sin 
kt or f 
t cos 
kt makes k complete cycles
between 0 and 2, and each functions has a period of:
Period=
Note: we are working in radians, else
360
k
2
k
The Period of tan(kt)
If k>0, then the graph of
and

f 
t tan 
kt makes k complete cycles between  0
2

, and each functions has a period of:
2
Period=
Note: we are working in radians, else

k
360
k
Example:
Determine the period of each of the functions:
a)
b)
c)
y sin 
3x 
t 
f 
t cos  
2 
3 
k 
 tan  
2 
Solutions:
a)
period =
2
3
We have 3 complete cycles between 0 and 2
b)
period =
2
4
1
2
½ cycle between 0 and 2
c)
period =
 2

3
3
2
Note: A calculator may not produce an accurate graph of trigonometric functions
with a large k value. For example the graph of
complete cycles between 0 and
showing this (try it).
f 
t sin 
50t has 50
2, but some calculators have problems
Graphing The Sine and Cosine Trigonometric Functions by Hand
When we are asked to graph trigonometric functions by hand, two types of questions
are usually presented to you:
The first being that you have full control. A blank piece of graph paper is provided
and you are to sketch your graph on the grid. You may be asked to one cycle or a set
number of cycles, so use logic to guide you.
It is best to use 12 blank spaces or a multiple of 12 blank spaces for your cycle. Use
logic to determine which is best to provide the number of required cycles.
With 12 spaces (or multiple of 12) you can easily find the 4 intervals (or quarters) that
will correspond to the
  3 
0, ,, ,2locations that would have a value of {-1, 0, 1}
2
 2

values of either the sine or cosine function.
The second being that the grid with its domain is already provided for you. Here you
have to conform to its restrictions when graphing the function. There is a bit more
work here, but a fun puzzle to solve.
Type 1: When you are given a blank grid.
These are the easiest to graph for you are given a blank grid and have full control, so
make it easy for yourself.
If you are given no domain restrictions and you need to draw one complete cycle.
Step 1: Determine the Period
Step 2: Let 12 spaces on the grid represent the period.
Step 3: Determine the ¼ period, ½ period, ¾ period values for your function (these
will be 3 spaces apart on your grid).
Step 4: Plot your sin or cos graph by using (0, 1, 0, -1, 0, 1, 0, -1) amplitude points
and follow the pattern.
If you need more than one cycle, see how many multiples of 12 you can use on your
grid. Use logic to assist you, as you will need to break each cycle in to quarters, so you
may have to use 8 grid spaces, or 4 grid spaces per cycle.
Some questions may require you to plot both positive and neagative domain values. If
this is the case ensure you provide a symmetric axis location.
Example:
Graph
f 
t sin 
2t when given a blank grid
Solution:
Determine the period:
2

2
Now horizontal grid markers will be at ¼ period, ½ period, ¾ period
(Q1)
1




4
4
1

Q 2) 


2
2
3
3
Q 3) 


4
4
Since this is a sine function. The f(t) values will (starting at t=0) follow the pattern {0,
1, 0, -1, 0} at the quarter period t-values.
Plot the following 5 points:


  3 
0,0
,  ,1
, ,0 
,  , 1
,
,0 



4 2 4


Now connect the dots to complete the graph.
Example:
Graph
 
y cos  x when given a blank grid.
10 
Solution:
Determine the Period:
2
10
2 20


10
Now horizontal grid markers will be at ¼ period, ½ period, ¾ period
Q1)
1

205
4
Q 2)
1

2010
2
Q 3)
3

2015
4
Use these values to label your grid.
Since this is a cosine function. The y-values will (starting at x=0) follow the pattern
{1, 0, -1, 0, 1} at the quarter period x-values. Plot the following 5 points:
, 5,0 
, 10, 1
, 15,0 
, 20,1
0,1

Type 2: When you are given a labelled grid.
These are harder to do; but by following the technique below you will find these are
actually fun to plot. Remember the teacher or textbook will provide questions that fit
nicely into the grid that they provide. This makes your job very easy.
If possible, try to provide 12 blank spaces for the cycle (12 is divisible by 2, 3, 4, 6
and thus gives you more room to play). We will be using the 4 quarters of the cycle,
thus if 12 spaces for each cycle does not fit into the grid, try 8, or even 4. Again have
fun with the logic to make the graph fit into the set grid.
Step 1:
Determine blank space width by domain divided by 12.
Step 2:
Determine the period.
Step 3:
If your teacher set up the question correctly, the period should be a
multiple of blank grid squares. Count the numbers of grid squares for a
complete cycle. Use
period
.
blank square width
Step 4:
Take Step 4’s value and divide by 4, this new result will be the numbers of
squares each quarter of the function will take for the pattern (…, 0, 1, 0, 1, 0, 1, 0, …). Mark these quarter place markers on your grid starting at 0.
Step 5:
Place the (…, 0, 1, 0, -1, 0, 1, 0, …) at each marked of spot on your grid,
remember each quarter will be Step 4 number of spaces.
Step 6:
You should label your grid at the ¼ period, ½ period, ¾ period values for
your function.
Example:
Graph
y sin 
3x when domain is 0 x 2
Solution:
Our blank space width:
domain 2 
 
12
12 6
The period of the graph is:
2
3
The number of squares the period occupies is:
2
period
2 6
 3   4
1 black square width 
3 
6
The number of spaces per quarter is
4
1 .
4
Therefore every 4 grid spaces the function will complete a full cycle, with 1 space
providing the quarter function point at the {… 0, 1, 0, -1, 0, …} pattern.
This gives points at:

2 1 2 2 2 3
2 4 
0,0
,   ,1
,
 ,0 
,   , 1 
,
 ,0 

3 4 3 4 3 4
3 4 

Let’s graph it demonstrating the k=3 complete cycles.
Example
x 
y cos  when domain is x 2 and the grid is provided as below.
3 
Solution:
Since we have 60 spaces on the horizontal axis, and we would prefer 12, we will use 5
spaces per major tick.
Our blank space width :
domain 3 
 
12
12 4
The period of the graph is:
2
6
1
3
The number of squares a complete cycle occupies is:
period
6
4
 6 24

space width

4
Therefore every 24 grid spaces (with the grid only providing 12 spaces) the function
will complete a cycle with
24
6 spaces providing the quarter function point interval
4
for the {… 0, 1, 0, -1, 0, …} pattern.
In this question we will only be able to graph 1 point (the ¼ period value).
Since this is a cosine function, start at x=0 with (0,1) then its next point will be 6 grid
   3 
,0  ,0 .
 4  2 
space from 0 at the location 
6
Let’s graph it.
Example
Graph
15 
3
3
f 
 sin 
 for   on the following grid.
2
2
6 
Solution:
The grid spacing does not look nice. We have 18 major horizontal tick marks and
5 space-per-tick mark (giving a total of 90 horizontal spaces). Let’s first find the
period of our graph.
Period =
2
6 12 4
2 

15
15 15
5
6
Now since we have 18 horizontal ticks over a domain of 3, the width of each
horizontal tick is
3 
 .
18 6
Oh no, it looks like we have a problem. The period of
the horizontal tick spacing of
4
does not work well with
5

. We could use our calculator to approximate the
6
locations, or remember that the question is set up to work on the given grid
(most times). What about those 5 space-per-tick mark for a total of 90 spaces,
can we use them?
The width of each space is
3 

90 30
Now how many space does it take for the period of
4
?
5
4
period
4 30
Number of spaces per cycle =
 5   24
space width 
5 
30
Hey, this is a nice number to work with.
Each quarter cycle has a length of
24
6 spaces.
4
Since this is a Sine Function, it starts at (0,0) and each 6 spaces will follow the
usual Sine pattern. We can use the horizontal tick width of
the graph for reference to improve readability.

to place labels on
6