6.6: More Trigonometric Graphs
E. Kim
MTH 151
All notation and terminology is based on Swokowski, Cole. Algebra and
Trigonometry: with analytic geometry. Classic 12th Edition.
1
1
1
Goal
Last time we graphed
y = a sin(bx + c)
y = a cos(bx + c)
2
2
2
Goal
Last time we graphed
y = a sin(bx + c)
y = a cos(bx + c)
Build the graphs of
y = a csc(bx + c)
y = a sec(bx + c)
y = a tan(bx + c)
y = a cot(bx + c)
step-by-step
2
2
2
Review of Section 6.3
3
3
3
y = sin x and y = csc x
Reciprocal identity csc x =
1
sin x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sin(x)
4
4
4
y = sin x and y = csc x
Reciprocal identity csc x =
1
sin x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sin(x)
4
4
y = csc(x)
4
y = sin x and y = csc x
Reciprocal identity csc x =
1
sin x
3
2
period
1
−π
π
−1
2π
3π
−2
−3
y = sin(x) y = csc(x)
both functions have period 2π
4
4
4
y = cos x and y = sec x
Reciprocal identity sec x =
1
cos x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cos(x)
5
5
5
y = cos x and y = sec x
Reciprocal identity sec x =
1
cos x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cos(x)
5
5
y = sec(x)
5
y = cos x and y = sec x
Reciprocal identity sec x =
1
cos x
3
2
period
1
−π
π
−1
2π
3π
−2
−3
y = cos(x) y = sec(x)
both functions have period 2π
5
5
5
Use y = sin x and y = cos x to get y = tan x
tan x =
sin x
cos x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sin(x)
6
y = cos(x)
6
6
Use y = sin x and y = cos x to get y = tan x
tan x =
sin x
cos x
3
2
1
−π
π
−1
2π
7.5
3π
−2
−3
y = sin(x)
6
y = cos(x)
6
6
Use y = sin x and y = cos x to get y = tan x
tan x =
sin x
cos x
3
2
1
−π
π
−1
2π
7.5
3π
−2
−3
y = sin(x)
6
y = cos(x)
6
6
Use y = sin x and y = cos x to get y = tan x
tan x =
sin x
cos x
3
2
1
−π
π
−1
2π
7.5
3π
−2
−3
y = sin(x)
6
y = cos(x)
6
6
Use y = sin x and y = cos x to get y = tan x
tan x =
sin x
cos x
3
2
1
−π
π
−1
2π
7.5
3π
−2
−3
y = sin(x)
6
y = cos(x)
6
6
Use y = sin x and y = cos x to get y = tan x
tan x =
sin x
cos x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sin(x)
6
y = cos(x)
6
y = tan(x)
6
Use y = sin x and y = cos x to get y = tan x
tan x =
sin x
cos x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
6
6
6
Use y = sin x and y = cos x to get y = tan x
tan x =
sin x
cos x
period is π
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
period of tan x is π
6
6
6
Use y = sin x and y = cos x to get y = cot x
cot x =
cos x
sin x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sin(x)
7
y = cos(x)
7
7
Use y = sin x and y = cos x to get y = cot x
cot x =
cos x
sin x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sin(x)
7
y = cos(x)
7
y = cot(x)
7
Use y = sin x and y = cos x to get y = cot x
cot x =
cos x
sin x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cot(x)
7
7
7
Use y = sin x and y = cos x to get y = cot x
cot x =
cos x
sin x
period is π
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cot(x)
period of cot x is π
7
7
7
In the graphs of y = csc x, y = sec x, y = tan x, y = cot x
There is no “biggest” or “smallest” y value occuring:
8
8
8
In the graphs of y = csc x, y = sec x, y = tan x, y = cot x
There is no “biggest” or “smallest” y value occuring:
No notion of amplitude
8
8
8
In the graphs of y = csc x, y = sec x, y = tan x, y = cot x
There is no “biggest” or “smallest” y value occuring:
No notion of amplitude (contrast: y = sin x or y = cos x)
8
8
8
Goal
Build the graphs of
y = a csc(bx + c)
y = a sec(bx + c)
y = a tan(bx + c)
y = a cot(bx + c)
step-by-step
9
9
9
How about just a?
y = a csc x
y = a sec x
y = a tan x
y = a cot x
10
10
10
y = tan x vs y = 2 tan x vs y = 3 tan x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
11
11
11
y = tan x vs y = 2 tan x vs y = 3 tan x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
11
y = 2 tan(x)
11
11
y = tan x vs y = 2 tan x vs y = 3 tan x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
11
y = 2 tan(x)
11
y = 3 tan(x)
11
y = tan x vs y = − tan x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
12
12
12
y = tan x vs y = − tan x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
12
12
y = − tan(x)
12
y = tan x vs y = 21 tan x vs y = 13 tan x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
13
13
13
y = tan x vs y = 21 tan x vs y = 13 tan x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
13
y=
13
1
2
tan(x)
13
y = tan x vs y = 21 tan x vs y = 13 tan x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
13
y=
13
1
2
tan(x)
y=
1
3
tan(x)
13
y = sec x vs y = 2 sec x vs y = 3 sec x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
14
14
14
y = sec x vs y = 2 sec x vs y = 3 sec x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
14
y = 2 sec(x)
14
14
y = sec x vs y = 2 sec x vs y = 3 sec x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
14
y = 2 sec(x)
14
y = 3 sec(x)
14
y = sec x vs y = − sec x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
15
15
15
y = sec x vs y = − sec x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
15
15
y = − sec(x)
15
y = sec x vs y = 12 sec x vs y = 13 sec x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
16
16
16
y = sec x vs y = 12 sec x vs y = 13 sec x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
16
y=
16
1
2
sec(x)
16
y = sec x vs y = 12 sec x vs y = 13 sec x
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
16
y=
16
1
2
sec(x)
y=
1
3
sec(x)
16
What does a do?
Sign of a:
17
I
a > 0 ! don’t flip
I
a < 0 ! flip vertically
17
17
What does a do?
Sign of a:
I
a > 0 ! don’t flip
I
a < 0 ! flip vertically
Size of |a|:
17
I
|a| > 1 ! vertical expand
I
|a| < 1 ! vertical contract
17
17
How about just b?
y = csc bx
y = sec bx
y = tan bx
y = cot bx
18
18
18
y = tan bx
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
19
19
19
y = tan bx
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
19
y = tan(2x)
19
19
y = tan bx
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
19
y = tan(2x)
19
19
y = tan bx
3
smaller period
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
19
y = tan(2x)
19
19
y = tan bx
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
19
y = tan(2x)
19
y = tan(3x)
19
y = tan bx where b is a fraction
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
20
20
20
y = tan bx where b is a fraction
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
20
y = tan( 12 x)
20
20
y = tan bx where b is a fraction
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
20
y = tan( 12 x)
20
y = tan( 13 x)
20
b in y = tan(bx) or y = cot(bx)
What does b do?
21
I
Since original period of tangent and cotangent is π
I
For y = tan(bx) or y = cot(bx), period is
21
π
|b|
21
y = sec bx
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
22
22
22
y = sec bx
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
22
y = sec(2x)
22
22
y = sec bx
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
22
y = sec(2x)
22
22
y = sec bx
3
2
1
smaller
period
−π
π
−1
2π
3π
−2
−3
y = sec(x)
22
y = sec(2x)
22
22
y = sec bx
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
22
y = sec(2x)
22
y = sec(3x)
22
y = sec bx where b is a fraction
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
23
23
23
y = sec bx where b is a fraction
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
23
y = sec( 21 x)
23
23
y = sec bx where b is a fraction
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
23
y = sec( 21 x)
23
y = sec( 13 x)
23
b in y = csc(bx) or y = sec(bx)
What does b do?
24
I
Since original period of cosecant and secant is 2π
I
For y = csc(bx) or y = sec(bx), period is
24
2π
|b|
24
b and c together
y = tan(bx + c)
y = cot(bx + c)
y = csc(bx + c)
y = sec(bx + c)
25
25
25
How to graph y = tan(bx + c)
26
26
26
How to graph y = tan(bx + c)
Factor b
26
h c i
y = tan b x +
b
26
26
How to graph y = tan(bx + c)
Factor b
26
h c i
y = tan b x +
b
π
|b|
I
period =
I
phase shift = − cb
26
26
How to graph y = tan(bx + c)
Factor b
h c i
y = tan b x +
b
π
|b|
I
period =
I
phase shift = − cb
Consecutive vertical asymptotes for one branch of y = tan(bx + c)
found by solving
π
π
− < bx + c <
2
2
26
26
26
How to graph y = tan(bx + c)
Factor b
h c i
y = tan b x +
b
π
|b|
I
period =
I
phase shift = − cb
Consecutive vertical asymptotes for one branch of y = tan(bx + c)
found by solving
π
π
− < bx + c <
2
2
−
26
π
π
− c < bx < − c
2
2
26
26
How to graph y = tan(bx + c)
Factor b
h c i
y = tan b x +
b
π
|b|
I
period =
I
phase shift = − cb
Consecutive vertical asymptotes for one branch of y = tan(bx + c)
found by solving
π
π
− < bx + c <
2
2
−
π
π
− c < bx < − c
2
2
− π2 − c
<x<
b
26
26
π
2
−c
b
26
How to graph y = tan(bx + c)
Factor b
h c i
y = tan b x +
b
π
|b|
I
period =
I
phase shift = − cb
Consecutive vertical asymptotes for one branch of y = tan(bx + c)
found by solving
π
π
− < bx + c <
2
2
−
π
π
− c < bx < − c
2
2
− π2 − c
<x<
b
A branch takes up the interval x ∈ (
26
26
π
2
−c
b
− π2 −c
b ,
π
−c
2
b
)
26
Example: y = tan(2x + π2 )
Rewrite y = tan(2x + π2 ) as
h π i
y = tan 2 x +
4
27
27
27
Example: y = tan(2x + π2 )
Rewrite y = tan(2x + π2 ) as
h π i
y = tan 2 x +
4
27
π
|b|
I
period =
I
phase shift =
π
2
− cb
=
= − π4
27
27
Example: y = tan(2x + π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
28
28
28
Example: y = tan(2x + π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
28
y = tan(2x)
28
28
Example: y = tan(2x + π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
28
y = tan(2x)
28
y = tan(2x + π2 )
28
Example: y = 31 tan(2x + π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
29
29
29
Example: y = 31 tan(2x + π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
29
y = tan(2x)
29
29
Example: y = 31 tan(2x + π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
29
y = tan(2x)
29
y = tan(2x + π2 )
29
Example: y = 31 tan(2x + π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = tan(x)
29
y = tan(2x) y = tan(2x + π2 )
y = 13 tan(2x + π2 )
29
29
How to graph y = cot(bx + c)
30
30
30
How to graph y = cot(bx + c)
Factor b
30
h c i
y = cot b x +
b
30
30
How to graph y = cot(bx + c)
Factor b
30
h c i
y = cot b x +
b
π
|b|
I
period =
I
phase shift = − cb
30
30
How to graph y = cot(bx + c)
Factor b
h c i
y = cot b x +
b
π
|b|
I
period =
I
phase shift = − cb
Consecutive vertical asymptotes for one branch of y = cot(bx + c)
found by solving
0 < bx + c < π
30
30
30
How to graph y = cot(bx + c)
Factor b
h c i
y = cot b x +
b
π
|b|
I
period =
I
phase shift = − cb
Consecutive vertical asymptotes for one branch of y = cot(bx + c)
found by solving
0 < bx + c < π
−c < bx < π − c
30
30
30
How to graph y = cot(bx + c)
Factor b
h c i
y = cot b x +
b
π
|b|
I
period =
I
phase shift = − cb
Consecutive vertical asymptotes for one branch of y = cot(bx + c)
found by solving
0 < bx + c < π
−c < bx < π − c
−c
π−c
<x<
b
b
30
30
30
How to graph y = cot(bx + c)
Factor b
h c i
y = cot b x +
b
π
|b|
I
period =
I
phase shift = − cb
Consecutive vertical asymptotes for one branch of y = cot(bx + c)
found by solving
0 < bx + c < π
−c < bx < π − c
−c
π−c
<x<
b
b
π−c
A branch takes up the interval x ∈ ( −c
b , b )
30
30
30
Example: y = cot(2x − π2 )
Rewrite y = cot(2x − π2 ) as
π y = cot 2 x −
4
31
31
31
Example: y = cot(2x − π2 )
Rewrite y = cot(2x − π2 ) as
π y = cot 2 x −
4
31
π
|b|
I
period =
I
phase shift =
π
2
− cb
=
=
π
4
31
31
Example: y = cot(2x − π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cot(x)
32
32
32
Example: y = cot(2x − π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cot(x)
32
y = cot(2x)
32
32
Example: y = cot(2x − π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cot(x)
32
y = cot(2x)
32
y = cot(2x − π2 )
32
Example: y = 2 cot(2x − π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cot(x)
33
33
33
Example: y = 2 cot(2x − π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cot(x)
33
y = cot(2x)
33
33
Example: y = 2 cot(2x − π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cot(x)
33
y = cot(2x)
y = cot(2x − π2 )
33
33
Example: y = 2 cot(2x − π2 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cot(x)
33
y = cot(2x)
y = cot(2x − π2 )
33
y = 2 cot(2x − π2 )
33
How to graph y = sec(bx + c)
34
34
34
How to graph y = sec(bx + c)
Factor b
34
h c i
y = sec b x +
b
34
34
How to graph y = sec(bx + c)
Factor b
34
h c i
y = sec b x +
b
2π
|b|
I
period =
I
phase shift = − cb
34
34
Example: y = sec(x − π4 )
There is already a coefficient of 1 in front of x.
35
35
35
Example: y = sec(x − π4 )
There is already a coefficient of 1 in front of x. No factoring to do.
35
35
35
Example: y = sec(x − π4 )
There is already a coefficient of 1 in front of x. No factoring to do.
35
2π
|b|
I
period =
= 2π
I
phase shift = − cb =
π
4
35
35
Example: y = sec(x − π4 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
36
36
36
Example: y = sec(x − π4 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
y = sec(x − π4 )
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36
Example: y = 2 sec(x − π4 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
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37
37
Example: y = 2 sec(x − π4 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
y = sec(x − π4 )
37
37
37
Example: y = 2 sec(x − π4 )
3
2
1
−π
π
−1
2π
3π
−2
−3
y = sec(x)
y = sec(x − π4 )
y = 2 sec(x − π4 )
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37
37
How to graph y = csc(bx + c)
38
38
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How to graph y = csc(bx + c)
Factor b
38
h c i
y = csc b x +
b
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38
How to graph y = csc(bx + c)
Factor b
38
h c i
y = csc b x +
b
2π
|b|
I
period =
I
phase shift = − cb
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38
Example: y = csc(2x + π)
Rewrite y = csc(2x + π) as
π y = csc 2 x +
2
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39
39
Example: y = csc(2x + π)
Rewrite y = csc(2x + π) as
π y = csc 2 x +
2
39
2π
|b|
I
period =
I
phase shift =
2π
2 =π
− cb = − π2
=
39
39
Example: y = csc(2x + π)
3
2
1
−π
π
−1
2π
3π
−2
−3
y = csc(x)
40
40
40
Example: y = csc(2x + π)
3
2
1
−π
π
−1
2π
3π
−2
−3
y = csc(x)
y = csc(2x)
40
40
40
Example: y = csc(2x + π)
3
2
1
−π
π
−1
2π
3π
−2
−3
y = csc(x)
y = csc(2x) y = csc(2x + π)
40
40
40
Example: y = 21 csc(2x + π)
3
2
1
−π
π
−1
2π
3π
−2
−3
y = csc(x)
41
41
41
Example: y = 21 csc(2x + π)
3
2
1
−π
π
−1
2π
3π
−2
−3
y = csc(x)
41
y = csc(2x)
41
41
Example: y = 21 csc(2x + π)
3
2
1
−π
π
−1
2π
3π
−2
−3
y = csc(x)
41
y = csc(2x)
y = csc(2x + π)
41
41
Example: y = 21 csc(2x + π)
3
2
1
−π
π
−1
2π
3π
−2
−3
y = csc(2x)
41
y = csc(2x + π)
41
y=
1
2
csc(2x + π)
41
More graph transformations
42
42
42
y = | cos(x)| + 2
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cos(x), y = | cos(x)| and y = | cos(x)| + 2
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y = | cos(x)| + 2
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cos(x), y = | cos(x)| and y = | cos(x)| + 2
43
43
43
y = | cos(x)| + 2
3
2
1
−π
π
−1
2π
3π
−2
−3
y = cos(x), y = | cos(x)| and y = | cos(x)| + 2
43
43
43
Recall y = 2 sin(x)
3
2
1
−π
π
−1
2π
3π
−2
−3
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44
Recall y = 2 sin(x)
3
2
1
−π
π
−1
2π
3π
−2
−3
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44
What about y = f (x) sin(x)?
f (x) is damping factor
45
45
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What about y = f (x) sin(x)?
f (x) is damping factor
“damped sine wave”
y = f (x) sin(x)
45
45
45
Recall y = 2−x sin(x)
3
2
1
−π
π
−1
2π
3π
−2
−3
In gray are y = ±2−x
46
46
46
Recall y = 2−x sin(x)
3
2
1
−π
π
−1
2π
3π
−2
−3
In gray are y = ±2−x
46
46
46
Recall y = 2−x sin(x)
3
2
1
−π
π
−1
2π
3π
−2
−3
In gray are y = ±2−x
46
46
46
y = cos(x) + sin(x)
3
2
1
−π
π
−1
2π
3π
−2
−3
47
47
47
y = cos(x) + sin(x)
3
2
1
−π
π
−1
2π
3π
−2
−3
47
47
47
y = cos(x) + sin(x)
3
2
1
−π
π
−1
2π
3π
−2
−3
47
47
47
Just for fun: y = 2 cos(x) + .4 sin(5x) + .8 sin(8x)
3
2
1
−π
π
−1
2π
3π
−2
−3
See http://method-behind-the-music.com/mechanics/physics
How do we instantly recognize the sounds of various musical
instruments?
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48
Just for fun: y = 2 cos(x) + .4 sin(5x) + .8 sin(8x)
3
2
1
−π
π
−1
2π
3π
−2
−3
See http://method-behind-the-music.com/mechanics/physics
How do we instantly recognize the sounds of various musical
instruments?
48
48
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