Transcript Sec 2.3

Chapter 2
Trigonometric Functions of Real
Numbers
Section 2.3
Trigonometric Graphs
Trigonometric Graphs
The graphs of trigonometric functions are usually represent on a ty-coordinate
system. The t-axis runs horizontally and the y-axis runs vertically. The values for t
usually represent radians. This is because if degrees were used the graphs
themselves would look greatly distorted if you do not adjust the scaling.
Periodic Functions
Since all the trigonometric functions are combinations of the x and y coordinates of
points on the unit circle they will repeat every time you go around the unit circle.
sin(t) =sin(t+2)
and
cos(t) =cos(t+2)
This enables you to find the graph of sin t and the cos t between 0 and 2 then
just repeat the same pattern in the graph.
t
Terminal Point
sin t
cos t
0
(1,0)
0
1
/2
(0,1)
1
0

(-1,0)
0
-1
3/2
(0,-1)
-1
0
2
(1,0)
0
1
The graphs for each are shown below.
cos t
sin t
1
1
0.5
0.5
t
t
1
2
3
4
5
1
6
2
3
4
5
6
-0.5
-0.5
-1
-1
Both of the graphs above show one wave of the functions y = sin t and y = cos t.
If the graph continued it would look as follows:
-10
1
1
0.5
0.5
-5
5
10
-10
-5
5
-0.5
-0.5
-1
-1
10
Amplitude
How high above and below the waves centerline it goes is called the amplitude.
Putting a number (other than 1) in front of the sine and cosine will change its
amplitude by that number. A negative number in front flips it around the t-axis.
y  2 sin t
y  6 cos t
2
6
4
1
2
-6
-4
-2
2
4
6
-6
-4
-2
2
4
6
-2
-1
-4
-2
-6
0.4
4
0.2
-6
-4
-2
2
2
-0.2
-0.4
y
1
sin t
2
4
6
-6
-4
-2
2
4
-2
-4
y  5 cos t
6
Period and Phase Shift
The period of a trigonometric function is the distance along the t-axis it takes for
the graph to go through a complete cycle. The phase shift is where it begins
one of it cycles. The period and the phase shift are determined by the values for
b and c in the equations:
y = sin(bt+c)
-6
-4
and
y = cos(bt+c)
1
1
0.5
0.5
-2
2
4
6
-6
-4
-2
2
-0.5
-0.5
-1
-1


y  sin  3t  
6

1  
y  cos t  
4
2
4
6
To find the interval on which a function of the form y = sin(bt+c) or y = cos(bt+c)
will go through one of it cycles take the inside and set it equal to 0 and 2 and
solve.
Solving for t when the inside is set equal to 0 will give you the starting value of a
cycle (on the t-axis).
Solving for t when the inside is set equal to 2 will give you the ending value of a
cycle (on the t-axis).
Subtracting the values for the two solutions will give the period.
Example: y = sin(6t+)
1
beginning
6t    0
6t  

t
6
Phase Shift

6
ending
6t    2
6t  
t

0.5
-0.4
-0.2
0.2
-0.5
6
Period
   2 



6
6
6
3
-1
Graph between /6 and -/6
0.4
Displacement
The entire graph of y = sin(t) or y = cos(t) can be shifted up or down by adding a
number onto the end. This amount the graph gets moved up or down is called
the displacement. It is determined by the number d in the following equations:
y = sin(t) + d
and
y = cos(t) + d
If d is positive the graph is shifted up by d and d is negative the graph is shifted
down by d.
-6
-4
4
4
3
3
2
2
1
1
-2
2
4
6
-6
-4
-2
2
-1
-1
-2
-2
-3
-3
-4
-4
y = sin(t) + 2
y = cos(t) - 1
4
6
Putting it Together
The graphing information for the equations:
y = a sin(bt+c) +d
and
y = a cos(bt+c) + d
can be broken down into the following parts:
a : amplitude controls how far above and below the center line a cycle is
b and c: determine the period and phase shift which are where a cycle starts and
how long its is in the t direction
d: the displacement determine how far up or down the centerline of the graph is
moved.
y = 2 cos(3t+) – 1
amplitude = 2
displacement = -1

phase shift 
3
period 

3

  2

3
3
3t    0
3t  

t
3
3t    2
3t  

t
3
1
-1
-0.5
0.5
-1
-2
-3
1