Periodic Function Review
Key Topics
• Graphing periodic functions; amplitude, period, phase shift,
maximum, minimum, domain and range.
• Equation of periodic function from graphs.
max  min
max  min
a
,d 
2
2
2
k
period
c : read from the graph
• Word problems.
• Solving trig. Equations.
 

y  3sin  xx3 11
 3

Amplitude = 3
Period = 6
Phase shift = 3 to the right
Maximum y = 4
Minimum y =  2
3sin
3 2 x     
y  sin  2  x  1  1
4 4 
2 
3
Amplitude =
4
Period = 

Phase shift =
to the right
2
 3 7 
Maximum point =  , 
 4 4
 5 1 
Minimum point ==  , 
 4 4
y  3tan  2 x   1
Amplitude = N.A.
Period =

24
Phase shift = none.


2 2kk11  

 xx| x| x , x, x
, k, k 
Domain =
44


Range =  y y  
k
 

y  3sin  xx3 11
 3


3
2
period =
6
 
 
3
4
4
1
1
3
3
2
2
5
5
Cosecant Function
y  csc  x 
Domain:
{ x | x  R, x  k , k  Z }
Range:
{ y | y  R, y  1 or y  1}
Period: 2
V.A.: x  k , k is an integer.
Secant Function
Domain:
y  sec  x 
{ x | x  R, x  k   , k  Z }
2
Range:
{ y | y  R, y  1 or y  1}
Period: 2
V.A.: x  k   , k is an integer.
2
Tangent Function
Domain:
y  tan  x 
{ x | x  R, x  k   , k  Z }
2
Range:
{ y | y  R}
Period: 
V.A.: x  k   , k is an integer.
2
Cotangent Function
y  cot  x 
Domain:
{ x | x  R, x  k , k  Z }
Range:
{ y | y  R}
Period: 
V.A.: x  k , k is an integer.
Find Equation
Write two defining equations for the graph below. Give one sine
function and one cosine function.
Cosine function:
Period = 7
2
k
7
Max. = 1, min. =  5
a  3, d  2
No phase shift.
 2
y  3cos 
 7

x  2

Find Equation
Write two defining equations for the graph below. Give one sine
function and one cosine function.
Sine function:
Period = 7
2
k
7
Max. = 1, min. =  5
a  3, d  2
Shift 5.25 right.
 2

y  3sin   x  5.25    2
 7

Find Equation
 
Give a cosine function that has a maximum at  ,8  followed
4 
 9 
immediately by another maximum at  ,8  and has a
 4 
minimum value of 4.
Cosine function:
Period = 2 , k  1
Max. = 8, min. = 4
a  2, d  6
Shift

4
to the right.


y  2cos  x    6
4

Word Problem
A study is conducted and a certain population is observed. The
  t  2 
population is modelled by P(t )  20000  4000 cos  

12

 
where t is time in months from the beginning of the study.
• What is the minimum population? 16000
• What is the period of the population cycle? k 

,
12
period = 24 (months)
Word Problem
A study is conducted and a certain population is observed. The
  t  2 
population is modelled by P(t )  20000  4000 cos  

12

 
where t is time in months from the beginning of the study.
• At what month(s) during the first cycle does the population reach
22000?
  t  2 
1
 5
Let  =  
,
cos


,


,


2
3 3
  12  
t 2 
 , t 6

 12  3
 t  2  5


, t  22

 12  3

Word Problem
When throwing a ball, the horizontal distance travelled, d, in feet
v2
is modelled by the equation d  sin  2  , where v is velocity
32
with which the ball is thrown (in feet/s) and θ is the angle of
elevation (in radians). If the ball is thrown at a speed of 90 feet/s.
What angle of elevation is required for the ball to travel 170 feet?
902
170 
sin  2  , sin  2  0.67
32
2

0.74,   0.74
0.37, 1.20
2.41
The ball should be thrown at angle
of 0.37 or 1.20 radian.
Word Problem
At a certain ocean bay, the max. height of the water is 4 m above
mean sea level at 8:00 a.m. The height is at a max. again at 8:24
p.m. Create a sinusoidal function using height, h, in metres, and
time, t, in hours to model the situation. Use the model to
determine the height of the water above mean sea level at 10:00
a.m.
2
Period = 12.4, k 
12.4
a  4, d  0
 

h  t   4 cos 
 t  8 
 6 .2

 

h 10   4 cos 
10  8 
 6.2

2.12  m 
Trig. Equation
Solve 4cos 2  2 x   3 on the interval x   0, 2  . EXACT answers
Only.
3
3
Let  = 2 x, cos 2   , cos   
4
2
3
 3
cos  
cos 
2
2
 
5 5
2 x    n 2n
2 x     n 2n
12 6
12 6
or or
or or
11 11
7 7
x


n

2 
 2n
2 x     n 2n
12 6
12 6

13 11 23
x ,
,
,
12 12
12
12
5 17 7 19
x
,
,
,
12
12
12
12
Trig. Equation
2
6csc
x  5csc x  6  0 on the interval x   0, 2  . Round
Solve
answer to 2 decimal places.
 3csc x  2    2csc x  3  0
3
2
csc x  , sin x 
2
3
x 0.73
2
csc x 
3
or
csc x  1 or  1
x 0.73, 2.41
x   0.73 2.41
Trig. Equation
Solve (1  cos 2 x)(sin x  2)  7sin x  4 on the interval  2 , 2  .
Round answer to 2 decimal places.
(sin 2 x)(sin x  2)  7sin x  4
sin 3 x  2sin 2 x  7sin x  4  0
 sin x  1  sin 2 x  3sin x  4   0
 sin x  1  sin x  4   0
2
x
 3
2
,
2
Trig. Equation
cos x 1  sin x

 2 on the interval  2 , 2 .
Solve
1  sin x
cos x
cos x
1
Let
 y, y   2  0, y 2  2 y  1  0,
1  sin x
y
 y  1
2
0
cos x
 1, cos x  1  sin x, cos x  sin x  1 square both sides
1  sin x
cos 2 x  2cos x  sin x  sin 2 x  1, sin x  cos x  0
sin x  0, x  2 ,   , 0,  , 2
cos x  0