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