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Graphing the reciprocal trig functions.
csc , sec , coty x y x y x
Example 1. Graph csc , first graph sin , then use
reciprocals to find the graph of csc .
y x y x
y x
Here is the graph of siny x
reciprocalWe want the trig function
cscy x
1So we take , the reciprocal of all
sin
the trig values of sin
yx
y x
Sine function: 1 1, the only problem
with reciprocals here is when ? ? ? ? ? ?
y
1The values are equal to 0, is undefined and here we have asymptotes!
0y
Vertical Asymptotes!
At: 2 , ,0, , 2 etc.x
1 1
1 1 0.1 10
1 2 0.05 20
2
y yy y
1 1
1 1 0.1 10
1 2 0.05 20
2
y yy y
This gives us the graph:
NOTE: the closer we get to 0,
as 0, 1
y
y
y
We can do the same thing for the graph of: sec ,
that is graph cos first, then draw the vertical
asymptotes where cos 0 and sketch sec .
y x
y x
x y x
Graph cos to get:y x
Now for the asymptotes:Now use camel technique:
draw on the humps!
Now for coty x
Graph tan first, then draw the vertical
asymptotes where tan 0 and sketch cot .
y x
x y x
Now the vertical asymptotes appear
where the tangent function 0
Now where the tangent function
had asymptotes, the cot function 0
Now where the tan 1 or tan 1,
cot 1 or cot 1
x x
x x
Now join the points between the
asymptotes!
Properties of the reciprocal trig functions:
I. csc , Domain: , , , Integers
Range: 1 1, asymptotes: ,
y x x n n
y y x n n I
II. sec , Domain: , , , Integers2
Range: 1 1, asymptotes: , , odd2
ny x x n
ny y x n I
III. cot , Domain: , , , Integers
Range: , , asymptotes: , ,
y x x n n
x n n I
Suppose we were asked to graph: 2sec 2 1.y x
The center of the graph is: 1, where the graph crosses this
line is where the asymptotes occur!
y
2We would graph: 2cos 2 1, , ,
2 42, no phase shift, down 1
y x P I
A
The starting point would be: 0, 0,1 this is high point, then
right down 2, right down 2,right up 2,right up 2,4 4 4 4etc
a d
Now we can
join the points.
The starting point would be: 0, 0,1 this is high point, then
right down 2, right down 2,right up 2,right up 2,4 4 4 4etc
a d
Center is 1y
Now where the
graph crosses the
center line is
where our vertical
asymptotes occur!
Now we can use the camel technique to sketch the graph on the humps
of the cosine graph!
Graph: 2cot 2 1y x / , but a vertical stretch by a factor of 2, , I , PS none, 1 up
2 2 4
PA N A P VT
b
graph cot first, , I , Asymptotes: , , hence 0, , 2 , etc.2 2
Py x P y n n I y
Method I, transformational approach,
x 2x
x
0x Now some points:
,02
3 , 14
,14
Now let us consider: 2coty x
This is a vertical stretch by a factor of 2This is the previous graph, we now multiply
all the values by 2 to get:y
Now some points: , 24
,02
3 , 24
Asymptotes still the same!
0x x 2x
x
Now we are ready to tackle: 2cot 2 1y x
, I , this is horizontal shrink2 2 4
PP
b
1 1 3Asymptotes: , , , .
2 2 2x n etc And 1 unit up !!
Now we have the following graph: 2cot 2y x
To go to 2cot 2 1, we move
everything up 1 unit !
y x
SOME POINTS: ,14
3 , 18
,38
2x 0x x
2x
Consider Method II:
For 2cot 2 1, we know that the , ,2 4
therefore the asymptotes go from , to
1 3, which gives: 0, , , , . . .
2 2 2If we look at cot 2 we have:
y x P I
x n n I
x n x
y x
4,0
8,1
38, 1
Now we multiply all 's by 2 & add 1y ,38
,14
3,1 8
0x 2x 2x
NOW we have: 2cot 2 1 y x