Transcript 4.6 - bcarroll01

Graphs of other
Trig Functions
Section 4.6

What is the cosecant x?
1
Sin x

Where is cosecant not defined?
◦ Any place that the Sin x = 0
x = 0, π, 2 π

The curve will not pass through these
points on the x-axis.
Cosecant Curve

Drawing the cosecant curve
Draw the reciprocal curve
2) Add vertical asymptotes wherever curve
goes through horizontal axis
3) “Hills” become “Valleys” and
“Valleys” become “Hills”
1)
Cosecant Curve
→ y = Sin x
y = Csc x
1

2

-1
Cosecant Curve
3
2
2
y = 3 Csc (4x – π) → y = 3 Sin (4x – π)
c=π 
a=3 b=4 
Per. = 2
P.S. = 4
3
-3
dis. =

8

4
3
8
Cosecant Curve

2
5
8
3
4
y = -2 Csc 4x + 2 → y = -2 Sin 4x + 2
4
2

8

4
Cosecant Curve
3
8

2

What is the secant x?
1
Cos x
Where is secant not defined?
◦ Any place that the Cos x = 0
 3
2 2
 The curve will not pass through these
points on the x-axis.

Secant Curve
y = Sec 2x
→ y = Cos 2x
1
-1

4

2
Secant Curve
3
4

y = Sec x
→ y = Cos x
1
-1

2

Secant Curve
3
2
2
1)
2)
3)
4)
y = 3 Csc (πx – 2π)

y = 2 Sec (x + )
2

y = ½ Csc (x - )
4
y = -2 Sec (4x + 2π)
Graph these curves
y = 3Csc (πx – 2π) → y = 3 Sin (π x – 2π)
3
2
-3
5
2
3
7
2
4
y = 2Sec (x

+2
)
→ y = 2 Cos (x +

2
3
2
2


2
-2


)
2
y = ½ Csc (x -

)
4
→ y = ½ Csc (x -

4
)
½
-½

4
3
4
5
4
7
4
9
4
y = -2 Sec (4π x + 2 π)
-2 Cos (4π x + 2 π)
2
1

2
3

8
1

4
1

8
-2
Graph of Tangent and
Cotangent
Still section 4.6

Define tangent in terms of sine and cosine
Sin x
Cos x

Where is tangent undefined?
Wherever Cos x  0
x-
 
,
2 2
Tangent
 asymptotes
y = Tan x


2
0

2
So far, we have the curve and 3 key
points
 Last two key points come from the
midpoints between our asymptotes and
the midpoint



◦ Between  and 0 and between
and 0
2
2


→ 
and
4
4
Tangent Curve
y = Tan x




x
2
4
y =Tan x und. -1
1

 0 


-1
2
4
4
0

4

2
0
1
und.

2

For variations of the tangent curve
1)
Asymptotes are found by using:
A1. bx – c = 

2
A2.

bx – c =
2
A1  A2
2) Midpt. =
2
A1  Midpt
A2  Midpt
3) Key Pts:
and
2
2
y = 2Tan 2x


x
4
y =2Tan 2x und.
bx – c =
2x=
x=




4

2

2

4
und.
bx – c =
2x =
x=

2

2

4


4

4
y = 2Tan 2x




x
4
8
y =2Tan 2x und. -2

Midpt =
=0
0
4
2


4
=
0
2
0
2

4
und.

4
4

K.P. =

2

K.P. =

0

8
=


8
8


4


8
0

8

4
y = 4Tan
x
y =4Tan
x
2
x
2
2
0

2
und. -4
0
4



A2  Midpt A1  A2
A1  
Midpt

P. 
K bx
.P.c  
bx Kc.
Midpt 
2
2
2
2
x   0
x  0
K .P. 
 
K .P. 2   2
Midpt 
2
2
2
2
2
 

x


x

K .P.  
K .P. 
Midpt  0
2
2

und.
y = 4Tan
x
y =4Tan
x
2
x
2
2
0

2
und. -4
0
4



4



2
0 
2
4


und.

Cotangent curve is very similar to the
tangent curve. Only difference is
asymptotes
bx – c = 0
bx – c = π
→ 0 and π are where Cot is undefined
Cotangent Curve
y = 2Cot
x

(x  )
2

2
3
4
π
2Cot ( x  2 ) und. 2
0

5
4
-2
A1  A2
Midpt
bx  cA10Midpt bx  c   A2 Midpt
K .P. 
2
K .P. 
2


 3
2
3

x 
x   0



2
2 
2 2
2
Midpt

K .P. 3
K .P.  x2 
2
2
x

22
2 5 Midpt  
3
K .P. 
K .P. 
4
4
3
2
und.
y = 2Cot
x
2Cot ( x 

(x  )
2

2

2
)
3
4
π
und. 2
0
2
2

2
3
4

5
4
5
4
-2
3
2
3
2
und.
y = 3 Cot
x
3Cot ( x 

(x  )
4

4

4
)

2
3
4

5
4
und. 3
0
-3
und.
3
3


4

2
3
4
5
4

Graph the following curves:
5
5 x
y = 2 Cos (
+
8
2
y = 2 Sin (
x
2
+π)+1
y = 5 Tan (4x – π )
)+2
y
a
5
5 x
= 2 Cos ( 2 + 8 ) +
5
5
=2 b= 2
c= 8
4
1
Per. = 5 P.S. = 
4
14
dis. = 5
2
d=
2
2
1

4
1

20
3
20
7
20
11
20
y = 2 Sin (

a=2 b= 2
Per. =
4
dis. =
1
x
2
+π)+1
c=

d=
P.S. =  2
3
1
2
1
-1
1
2
1
y = 5Tan (4x – π)

3
x
8
16
5Tan (4x – π) und. -5

8

4

4
5
16
0
5
3
8
3
8
und.

Graph the following curves:

y = -3 Sec (x +
)
2

y = -2 Csc (x )
4

y = ½ Cot (x – )
4
y = -3 Sec (x +

2
)
-3 Sec ( x

+2
3


2
-3

2

3
2
)
y = -2 Csc (x -

4
)→ y = -2 Csc (x

4
)
2
-2

4
3
4
5
4
7
4
9
4

y = ½ Cot( x  4 )

x
4
½ Cot( x 
1
2
1

2

4
3
4

2
und. ½
)

0
- ½ und.


4

2
3
4
5
4
5
4