SECTION THREE: GRAPHS OF TRIGONOMETRIC FUNCTIONS …jwilson.coe.uga.edu/EMAT6680/Caglayan/iu/i3.pdfSECTION THREE: GRAPHS OF TRIGONOMETRIC FUNCTIONS AND INVERSE TRIGONOMETRIC FUNCTIONS:

SECTION THREE: GRAPHS OF TRIGONOMETRIC FUNCTIONS AND INVERSETRIGONOMETRIC FUNCTIONS:

3.1 PERIODIC FUNCTIONS:

If f is a function such that f(x+T)=f(x) for a real number T, then f is said to be aperiodic function. T is called the period of f. If T is smallest positive such number,then it is called fundamental period of f. Even in that case, sometimes it is just calledperiod of f.

Example: Let f be a function f :ℤ{0,1,2,3,4} defined by the rule: f(x) is theremainder when x is divided by 5. For instance,

f(-1)=4, f(-1+5)=f(4)=4 f(0)=0, f(0+5)=f(5)=0f(-9)=1, f(-9+5)=f(-4)=f(-4+5)=f(1)=1

That is, for each x in the domain of f, we have f(x+5)=f(5)

So the same pattern repeats itself and the fundamental period is 5. From now on inthis text, by period we mean fundamental period.

3.2 PERIODS OF TRIGONOMETRIC FUNCTIONS:

1. SINE: Since sin xk⋅2=sin x for k∈ℤ , sine function is periodic and theperiod is 2 .

2. COSINE: Since cosxk⋅2=cos x for k∈ℤ , cosine function is periodic and theperiod is 2 .

3. TANGENT: Since tan xk⋅=tan x for k∈ℤ , tangent function is periodic andthe period is .

4. COTANGENT: Since cot xk⋅=cot x for k∈ℤ , cotangent function is periodicand the period is .

4

2

y

-10 -5 5

x

3.3 GRAPHS OF TRIGONOMETRIC FUNCTIONS:

1. GRAPH OF THE COSINE FUNCTION:

Instead of graphing cosine function by plotting points in a traditional way, we willuse the unit circle. Remember how we defined cosine in the unit circle. That's whatwe will do here. GSP will translate the unit circle to the cartesian coordinate system.Snapshots of this animation are given below. To download the GSP animation file,please go to the site.

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad0.00=rad= 0.00m AP = 0.26°

C: (0.00, 1.00)P: (1.00, 0.00)

SNAPSHOT # 1As point P rotates on the unit circle, Point C traces cosine function. If θ=m∠AOP, then point P has coordinates: P(cos θ ,sin θ)Animate Point P on the unit circle

C

O AP

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad0.17=rad= 0.52m AP = 29.92°

C: (0.52, 0.87)P: (0.87, 0.50)


C

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad0.25=rad= 0.79m AP = 45.37°

C: (0.79, 0.70)P: (0.70, 0.71)


C

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad0.33=rad= 1.05m AP = 59.90°

C: (1.05, 0.50)P: (0.50, 0.87)


C

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad0.50=rad= 1.57m AP = 90.00°

C: (1.57, 0.00)P: (0.00, 1.00)


CO A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad0.67=rad= 2.10m AP = 120.14°

C: (2.10, -0.50)P: (-0.50, 0.86)


CO A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad0.75=rad= 2.36m AP = 135.00°

C: (2.36, -0.71)P: (-0.71, 0.71)


C

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad0.83=rad=2.62m AP = 149.96°

C: (2.62, -0.87)P: (-0.87, 0.50)


C

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad1.00=rad=3.15m AP = 180.21°

C: (3.15, -1.00)P: (-1.00, 0.00)


C

O AP

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad1.17=rad=3.67m AP = 210.02°

C: (3.67, -0.87)P: (-0.87, -0.50)


C

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad1.25=rad=3.93m AP = 225.26°

C: (3.93, -0.70)P: (-0.70, -0.71)


C

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad1.33=rad=4.19m AP = 239.84°

C: (4.19, -0.50)P: (-0.50, -0.86)


CO A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad1.50=rad=4.72m AP = 270.26°

C: (4.72, 0.00)P: (0.00, -1.00)


CO A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad1.67=rad=5.24m AP = 300.19°

C: (5.24, 0.50)P: (0.50, -0.86)


C

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad1.75=rad=5.50m AP = 315.11°

C: (5.50, 0.71)P: (0.71, -0.71)


C

O A

P

Watching these snapshots one after one, it was like a movie was not it?

These snapshots will tell you everything about graphing cosine function. I am notgoing to explain how we graph functions by plotting point in a traditional way.Instead, download the GSP file from the site and examine the animation carefully.Let us finish this part by making a table that summarize all 17 snapshots: (Pleaserefer to the BIG unit circle from Section 2.9)

x 0 6

4

3

2

23

34

56

y=cos x 1 32

12

12

0 −12

− 12 −3

2

x 76

54

43

32

53

74

116

2

y=cos x −1−32

− 12

−12

0 12

12 −3

21

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad1.83=rad=5.76m AP = 330.24°

C: (5.76, 0.87)P: (0.87, -0.50)


C

O AP

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

π rad2.00=rad=6.28m AP = 359.90°

C: (6.28, 1.00)P: (1.00, 0.00)


C

O AP

2. GRAPH OF THE SINE FUNCTION:

Once again let's just use the unit circle. Remember how we defined sine in the unitcircle. That's what we will do here. GSP will translate the unit circle to the cartesiancoordinate system. Snapshots of this animation are given below. To download theGSP animation file, please go to the site.

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (0.01, 0.01)

π rad0.00=rad=0.01m AP = 0.36°P: (1.00, 0.01)

SNAPSHOT # 1As point P rotates on the unit circle, Point S traces sine function. If θ=m∠AOP, then point P has coordinates: P(cos θ ,sin θ)Animate Point P on the unit circle

SO AP

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (0.53, 0.51)

π rad0.17=rad=0.53m AP = 30.40°P: (0.86, 0.51)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (0.80, 0.71)

π rad0.25=rad=0.80m AP = 45.58°P: (0.70, 0.71)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (0.78, 0.71)

π rad0.25=rad=0.78m AP = 44.88°P: (0.71, 0.71)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (1.05, 0.87)

π rad0.33=rad=1.05m AP = 60.22°P: (0.50, 0.87)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (1.57, 1.00)

π rad0.50=rad=1.57m AP = 90.10°P: (0.00, 1.00)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (2.10, 0.86)

π rad0.67=rad=2.10m AP = 120.46°P: (-0.51, 0.86)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (2.36, 0.71)

π rad0.75=rad=2.36m AP = 135.05°P: (-0.71, 0.71)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (2.62, 0.50)

π rad0.83=rad=2.62m AP = 150.07°P: (-0.87, 0.50)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (3.15, 0.00)

π rad1.00=rad=3.15m AP = 180.27°P: (-1.00, 0.00)


SO AP

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (3.67, -0.50)

π rad1.17=rad=3.67m AP = 210.20°P: (-0.86, -0.50)


SO A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (3.93, -0.71)

π rad1.25=rad=3.93m AP = 225.16°P: (-0.71, -0.71)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (4.19, -0.87)

π rad1.33=rad=4.19m AP = 240.23°P: (-0.50, -0.87)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (4.71, -1.00)

π rad1.50=rad=4.71m AP = 269.84°P: (0.00, -1.00)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (5.23, -0.87)

π rad1.67=rad=5.23m AP = 299.88°P: (0.50, -0.87)


S

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (5.49, -0.71)

π rad1.75=rad=5.49m AP = 314.79°P: (0.70, -0.71)


S

O A

P

GSP did everything... Same as before, you can download the GSP file from the siteand examine the animation carefully. Let us finish this part by making a table thatsummarizes all 17 snapshots: (Please refer to the BIG unit circle from Section 2.9)

x 0 6

4

3

2

23

34

56

y=sin x 0 12

12

32

1 32

12

12

x 76

54

43

32

53

74

116

2

y=sin x 0 −12

− 12 −3

2−1

−32

− 12

−12

0

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (5.76, -0.50)

π rad1.83=rad=5.76m AP = 330.02°P: (0.87, -0.50)


SO A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

S: (6.28, -0.01)

π rad2.00=rad=6.28m AP = 359.57°P: (1.00, -0.01)


SO AP

3. GRAPH OF THE TANGENT FUNCTION:

Here again we will use the unit circle. Remember how we defined tangent in the unitcircle. That's what we will do here. GSP will translate the unit circle to the cartesiancoordinate system. Snapshots of this animation are given below. To download theGSP animation file, please go to the site.

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

T: (0.01, 0.01)

π rad0.00=rad=0.01m AP = 0.31°P: (1.00, 0.01)

SNAPSHOT # 1As point P rotates on the unit circle, Point T traces tangent function. If θ=m∠AOP, then point P has coordinates: P(cos θ ,sin θ)Animate Point P on the unit circle

TO A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

T: (0.53, 0.58)

π rad0.17=rad=0.53m AP = 30.29°P: (0.86, 0.50)


T

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

T: (0.79, 1.01)

π rad0.25=rad=0.79m AP = 45.18°P: (0.70, 0.71)


T

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6

x

T: (1.05, 1.74)

π rad0.33=rad=1.05m AP = 60.05°

P: (0.50, 0.87)

SNAPSHOT # 4

As point P rotates on the unit circle, Point T traces tangent function.

If θ=m∠AOP, then point P has coordinates: P(cos θ ,sin θ)

Animate Point P on the unit circle

T

O A

P

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6 x

NOTE: GSP has gone crazy here because tangent isundefined for 90° (Look at the coordinates of trace point T)

T: (1.57, 4.41⋅106)

π rad0.50=rad=1.57m AP = 89.99999°

P: (0.00, 1.00)

SNAPSHOT # 5


If θ=m∠AOP, then point P has coordinates: P(cos θ,sin θ)


O A

P

2

1.5

1

0.5

-0.5

-1

-1.5

-2

y

-1 1 2 3 4 5 6 x

NOTE: Trace point T is back

T: (2.10, -1.73)

π rad0.67=rad=2.10m AP = 120.09137°

P: (-0.50, 0.87)

SNAPSHOT # 6


If θ=m∠AOP, then point P has coordinates: P(cos θ,sin θ)


T

O A

P

2

1.5

1

0.5

-0.5

-1

-1.5

y

-1 1 2 3 4 5 6 x

T: (2.36, -1.00)

π rad0.75=rad=2.36m AP = 135.13388°

P: (-0.71, 0.71)

SNAPSHOT # 7




T

O A

P

2

1.5

1

0.5

-0.5

-1

-1.5

y

-1 1 2 3 4 5 6 x

T: (2.63, -0.57)

π rad0.84=rad=2.63m AP = 150.52914°

P: (-0.87, 0.49)

SNAPSHOT # 8




TO A

P

2

1.5

1

0.5

-0.5

-1

-1.5

y

-1 1 2 3 4 5 6 x

T: (3.14, 0.00)

π rad1.00=rad=3.14m AP = 179.85434°

P: (-1.00, 0.00)

SNAPSHOT # 9




TO A

P

2

1.5

1

0.5

-0.5

-1

-1.5

-2

y

-1 1 2 3 4 5 6 x

T: (3.67, 0.58)

π rad1.17=rad=3.67m AP = 210.17772°

P: (-0.86, -0.50)

SNAPSHOT # 10




T

O AP

2

1.5

1

0.5

-0.5

-1

-1.5

-2

y

-1 1 2 3 4 5 6 x

T: (3.93, 1.01)

π rad1.25=rad=3.93m AP = 225.35766°

P: (-0.70, -0.71)

SNAPSHOT # 11




T

O AP

2

1.5

1

0.5

-0.5

-1

-1.5

-2

y

-1 1 2 3 4 5 6 x

T: (4.19, 1.74)

π rad1.33=rad=4.19m AP = 240.05314°

P: (-0.50, -0.87)

SNAPSHOT # 12




T

O A

P

2

1.5

1

0.5

-0.5

-1

-1.5

-2

y

-1 1 2 3 4 5 6 x

Note : Tangent is undefined for 270°. (Trace point T vanished)

T: (4.71, 3253.41)

π rad1.50=rad=4.71m AP = 269.98239°

P: (0.00, -1.00)

SNAPSHOT # 13




O A

P

2

1.5

1

0.5

-0.5

-1

-1.5

-2

y

-1 1 2 3 4 5 6 x

Note: Trace point T is back

T: (5.23, -1.74)

π rad1.67=rad=5.23m AP = 299.85781°

P: (0.50, -0.87)

SNAPSHOT # 14




T

O A

P

2

1.5

1

0.5

-0.5

-1

-1.5

-2

y

-1 1 2 3 4 5 6 x


T: (5.50, -0.99)

π rad1.75=rad=5.50m AP = 315.36073°

P: (0.71, -0.70)

SNAPSHOT # 15




T

O AP

Let us finish this part by making a table that summarizes all 17 snapshots: (Pleaserefer to the BIG unit circle from Section 2.9)

x 0 6

4

3

2

23

34

56

y=tan x 0 13

1 3 ∞ −3 −1 − 13

x 76

54

43

32

53

74

116

2

y=tan x 0 13

1 3 ∞ −3 −1 − 13

0

2

1.5

1

0.5

-0.5

-1

-1.5

-2

y

-1 1 2 3 4 5 6 x


T: (5.76, -0.58)

π rad1.83=rad=5.76m AP = 329.89472°

P: (0.87, -0.50)

SNAPSHOT # 16




TO AP

2

1.5

1

0.5

-0.5

-1

-1.5

-2

y

-1 1 2 3 4 5 6 x

T: (6.28, 0.00)

π rad2.00=rad=6.28m AP = 359.72861°

P: (1.00, 0.00)

SNAPSHOT # 17




TO A

P

4. GRAPH OF THE COTANGENT FUNCTION:

Cotangent is defined as 1 over tangent. Therefore, here I will not repeat all thesnapshots. You can download the GSP animation file from the site. Here is the graphof cotangent traced by GSP:

3.4 INVERSE TRIGONOMETRIC FUNCTIONS AND THEIR GRAPHS:

Review: Graphs of functions f(x) and f^-1 (x) are symmetric with respect to liney=x.

Notation: To express the inverse of a trigonometric function, we put “arc” in front ofthe trigonometric function as a prefix. For example, for the inverse of cos x, we writearccos x, for the inverse of sin x, we write arcsin x, etc.

Horizontal Line Test: Horizontal lines intersect graphs of trigonometric functions atmore than one point. Therefore,trigonometric functions are not invertible:

To obtain a well-defined inverse, we restrict the domain and the range of thoseinverse trigonometric functions as follows:

arcsin x :[−1,1][−2,2] arccos x :[−1,1][0 ,]

arctan x :[−∞ ,∞][−2,2] arccot x : [−∞ ,∞][0 ,]

2

1.5

1

0.5

-0.5

-1

-1.5

y

-1 1 2 3 4 5 6 x

J: (2.03, -0.50)

π rad0.65=rad=2.03m AP = 116.52985°

P: (-0.45, 0.89)

COTANGENT

As point P rotates on the unit circle, Point J traces cotangent function.



JO A

P

1

-1

-5 5

f x( ) = sin x( )1

-1

-5 5

g x( ) = cos x( )1

-1

-5 5

h x( ) = tan x( )

1. ARCSINE FUNCTION:

Let us choose one of the intervals, say [−2,2] in which the sine function is

invertible. Hence the inverse of the function f :[−2,2][−1,1] , f x =sin x is

also a function. In this case we write f −1 : [−1,1][−2,2] , f −1x =arcsin x .

By using this definition, y= f x=sin x⇔ x= f −1 y =sin−1 y =arcsin y

Graph of y = arcsin x and the graph of y = sin x are symmetric to each other withrespect to line y=x. (Please go to the site to download the GSP file)

Exercise: Find for =arcsin 32 and =arcsin−2

2 provided that

∈[−2,2] and ∈[−

2,2]

6

5

4

3

2

1

-1

y

2 4 6 8

x

y=x

sin x and arcsin x are symmetricwith respect to line y=x.

y=arcsin x

y=sin x

S': (0.98, 1.38)S: (1.38, 0.98)

π rad0.44=rad=1.38m AP = 79.29°P: (0.19, 0.98)

As point P rotates on the unit circle, Point S traces sine function,Point S' traces arcsine function.

If θ=m∠AOP, then point P has coordinates: P(cos θ ,sin θ)Animate Point P on the unit circle

S'

S

O A

P

2. ARCCOSINE FUNCTION:

Let us choose one of the intervals, say [0 ,] in which the cosine function isinvertible. Hence the inverse of the function f :[0 ,][−1,1] , f x=cos x is also afunction. In this case we write f −1 :[−1,1][0 ,] , f −1x=arccos x .

By using this definition, y= f x=cos x⇔ x= f −1 y =cos−1 y =arccos y

Graph of y = arccos x and the graph of y = cos x are symmetric to each other withrespect to line y=x. (Please go to the site to download the GSP file)

Exercise: Find sin for =arccos 12,=arcsin−1

2,=arccos 3

2 provided that

, ,∈[0,]

5

4.5

4

3.5

3

2.5

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 6 7 8

x

y=x C': (-0.54, 2.14)

As point P rotates on the unit circle, Point C traces cosine function,Point C' traces arccosine function.

cos x and arccos x are symmetricwith respect to line y=x.

C: (2.14, -0.54)

P: (-0.54, 0.84)

As point P rotates on the unit circle, Point C traces cosine function. If θ=m∠AOP, then point P has coordinates: P(cos θ ,sin θ)

m∠AOP = 122.85°Animate Point P on the unit circle

C'

CO A

P

3. ARCTANGENT FUNCTION:

Let us choose one of the intervals, say [−2,2] in which the tangent function is

invertible. Hence the inverse of the function f : [−2,2]−∞ ,∞ , f x=tan x is also

a function. In this case we write f −1 : [−∞ ,∞][−2,2] , f −1x =arctan x .

By using this definition, y= f x=tan x⇔ x= f −1 y =tan−1 y =arctan y

Graph of y = arctan x and the graph of y = tan x are symmetric to each other withrespect to line y=x. (Please go to the site to download the GSP file)

Exercise: Determine arctan 3arctan−1 for the angles satisfying −2

2

4.5

4

3.5

3

2.5

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 x

y=x

T': (346.47, 1.57)T: (1.57, 346.47)

π rad0.50=rad=1.57m AP = 89.83463°

P: (0.00, 1.00)

As point P rotates on the unit circle, Point T traces tangent function,Point T' traces arctangent function.

tan x and arctan x are symmetricwith respect to line y=x.



O A

P

4. ARCCOTANGENT FUNCTION:

Let us choose one of the intervals, say [0 ,] in which the cotangent function isinvertible. Hence the inverse of the function f :[0 ,][−∞ ,∞] , f x =cot x is also afunction. In this case we write f −1 :[−∞ ,∞][0 ,] , f −1x=arccot x .

By using this definition, y= f x=cot x⇔ x= f −1 y=cot−1 y=arccot y

Graph of y = arccot x and the graph of y = cot x are symmetric to each other withrespect to line y=x. (Please go to the site to download the GSP file)

Exercise: Evaluate tan arccot 33cosarccot −3 for the angles satisfying 0

Exercise: Simplify sec arctan 1x

4.5

4

3.5

3

2.5

2

1.5

1

0.5

-0.5

-1

y

-1 1 2 3 4 5 x

y=x

x

J': (0.02, 1.55)

As point P rotates on the unit circle, Point J traces cotangent function,Point J' traces arccotangent function.

cot x and arccot x are symmetricwith respect to line y=x.

J: (1.55, 0.02)

π rad0.49=rad=1.55m AP = 88.71885°

P: (0.02, 1.00)



J'

JO A

P

Documents

SECTION THREE: GRAPHS OF TRIGONOMETRIC FUNCTIONS …jwilson.coe.uga.edu/EMAT6680/Caglayan/iu/i3.pdfSECTION THREE: GRAPHS OF TRIGONOMETRIC FUNCTIONS AND INVERSE TRIGONOMETRIC FUNCTIONS: