Circular Functions Versus Hyperbolic Functions

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3,2,4,5,3,1,2,4,3,3,2,1,2,3,1,3,2,3,1,5,6,1,2,1,2,3,1,3

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,4,3,2,3,1,2,3,1,3,3,2,3,3,4,3,2,1,5,1,5,1,2,1,5,1,3,2,

1,5,3,1,3,2,1,3,2,3,4,3,2,1,6,5,3,1,2,3,1,6,2,1,2,4,3,2 ,1,2,1,5,1,5,3,1,2,3,1,3,2,1,5,3,1,3,2,6,3,4,3,2,1,2,11. Circular Functions versus 4, 3,2,3,1,2,3,4,3,3,2,3,1,3,2,1,2,1,5,6,1,2,6,1,3,2,1,2,3Hyperbolic Functions ,3,1,6,3,2,9,1,2,1,2,4,3,2,3,1,2,4,3,3,2,1,2,3,1,3,2,1, 2,6,1,6,3,2,3,1,3,2,3,3,3,1,3,2,1,3,2,3,4,2,1,2,1,2,7,2 ,3,1,5,1,3,3,2,1,5,1,5,1,2,7,5,1,2,1,2,3,1,3,5,3,3,1,5,

1,3,2,3,4,2,1,2,3,4,3,5,1,2,3,1,3, 3,2,1,2,3,1,3,2,1,3,5

,1,5,3,1,2,3,4,2,1,2,6,1,3,2,1,3,2,3,6,1,2,1,2,4,3,2,3,

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1,2,4,3,2,3,1,2,3,1,3,3,3,2,3,4,2,1,2,1,5,6,1,2,1,5,1,3 ,2,1,2,3,3,1,5,1,3,2,7,3,2,1,2,4,5,3,1,2,3,1,3,3,2,1,2,"The geometry... is honoured with the title of the Key of the Sciences; ... it is the great master-key which unlock every door of knowledge and 4,3,2,1,2,6,1,6,2,1,2,3,1,3,2,1,without which no discovery – no discovery which5,3,1,3,2,4,2,3,4,2,1,2 deserves the name, which is law and no isolated fact – has been or ever can be made.” ,1,2,7,2,3,1,5,4,3,2,1,2,3,1,5,1,2,1,6,5,1,2,3,3,1,3,2,Professor Benjamin Peirce – President of the 3,3,3,1,3,3,3,2,3,6,1,2,3,4,3,5,1,2,4,3,3,2,1,2,3,1,3,2American Association for the year 1853 ,1,3,5,1,5,1,3,2,3,4,2,3,6,1,3,2,1,3,2,3,6,1,2,1,2,7,2, 3,1,2,3,1,3,5,1,5,1,3,2,1,2,6,1,5,1,2,3,3,1,3,3,2,3,3,1 ,5,1,3,2,3,4,2,1,3,2,4,3,2,3,1,2,3,4,3,2,1,5,1,3,2,1,3, 5,1,5,3,1,2,4,3,2,1,2,3,3,1,3,2,4,2,3,4,3,4,2,1,2,3,2,3 Complex Arithmetic – Patterns Behind Numbers

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1. The Circle and the Hyperbola

The connection between the circle and the hyperbola was first observed in the conical sections.

Figure 11-01: The circular and the hyperbolic angles

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2. Circular Angles – Circular Radians

Part of this section and the next section are based on Arthur Edwin Kennellyi book "The application of hyperbolic functions to electrical engineering problems"

Figure 11-02: Circular Sector and Circular Functions

As the radius OM rotates about the center O, the x diminishes from +1 to 푥푀, y increases from 0 to 푦푀, and the radius moves from OA to OM, going through c1, c2, c3; it describes a circular sector OAM and a circular angle ∝= 퐴푂푀. At any position M, the following lines (axis) are orthogonal:

푥 ⊥ 푦

푀퐶1 ⊥ 푦

푀퐶2 ⊥ 푥

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퐴퐶3 ⊥ 푥

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퐵퐶4 ⊥ 푦

퐶5퐶6 ⊥ 푂푀 At any point during the rotation the following equation holds:

푦2 + 푥2 = 1

The magnitude of the circular angle ∝ can be defined in two ways:

1. By the ratio of the circular arc length s described during the motion by the point M, to the length of the radius 푶푴 = 흆;

Let’s consider the infinitesimal element of the circular arc s as being defined by:

푑푠 = √(푑푥)2 + (푑푦)2 and consider the instantaneous value of radius of the circle length defined by:

휌 = √푥2 + 푦2 With these notations the element of the angle described during the rotation will be:

푑푠 푑훼 = − 푐𝑖푟푐푢푙푎푟 푟푎푑𝑖푎푛 휌

This is the infinitesimal element of the circular angle 훼 shown in Figure 1 and it will be expressed in units of circular radians (we will come back and define the circular radian).

As the circular motion proceeds, from an initial to a final position of the radius OM, the total angle described during the motion is:

푀 푑푠 훼 = ∫ 퐴 휌 If the radius of the circle is equal by assumption to unity the above equation becomes:

푀 푑푠 훼 = ∫ = 푠 − 푐𝑖푟푐푢푙푎푟 푟푎푑𝑖푎푛푠 퐴 1 Where s is the length of the circular arc AM, and 훼 is the corresponding circular angle expressed in circular radians (see Figure 11-3).

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Figure 11-03: A circular angle equal of one circular radian expressed by the ratio of the circular arc length to the radius. If OA = 1 then also AM = 1.

2. By the area of the circular sector AOM swept out by the radius during the motion from A to M.

In the circular sector defined in the Figure 11-02, the magnitude of the angle described by the radius OM, between the initial position A and the final position M, is numerically twice the area of the sector swept out by the radius during the rotation. Thus, if in Figure 11-3 the radius is equal with unity and the radius describes the circular arc Ac1c2c3M, then the circular angle, expressed in circular radians will be double the sector area of the sector OAM. This value is equal with the area of the sector OMM’ - see Figure 11-04.

In Figure 11-02 the circular angle 훼 is represented as 1 circular radian. It has to be observed that:

푂퐴 = 푂푀 = 퐴푀 = 1 − 푙푒푛𝑔푡ℎ 푢푛𝑖푡

퐴푟푒푎(푂푀퐴푀′) = 1 − 푎푟푒푎 푢푛𝑖푡

These two definitions are interchangeable, but for reasons that will become clearer as we move along in our investigation, we prefer and recommend the second definition. 248

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If, in Figure 11-02, 푂푀 = 푂퐴 = 푂퐵 = 1, the circular functions are defined as:

푂퐶1 = sin 훼

푂퐶2 = cos 훼

푂퐶3 = tan 훼

푂퐶4 = cot 훼

푂퐶5 = sec 훼

푂퐶6 = csc 훼 In Figure 11-02 the circular angle 훼 is represented as 1 circular radian. It has to be observed that:

푂퐴 = 푂푀 = 퐴푀 = 1 − 푙푒푛𝑔푡ℎ 푢푛𝑖푡

퐴푟푒푎(푂푀퐴푀′) = 1 − 푎푟푒푎 푢푛𝑖푡 These two definitions are interchangeable, but for reasons that will become clearer as we move along in our investigation, we prefer and recommend the second definition.

If, in Figure 11-02, 푂푀 = 푂퐴 = 푂퐵 = 1, the circular functions are defined as:

푂퐶1 = sin 훼

푂퐶2 = cos 훼

푂퐶3 = tan 훼

푂퐶4 = cot 훼

푂퐶5 = sec 훼

푂퐶6 = csc 훼 In Figure 11-02 the circular angle 훼 is represented as 1 circular radian. It has to be observed that:

푂퐴 = 푂푀 = 퐴푀 = 1 − 푙푒푛𝑔푡ℎ 푢푛𝑖푡

If, in Figure 10-02 푂푀 = 푂퐴 = 푂퐵 = 1, the circular functions are defined as: 249

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푂퐶1 = sin 훼

푂퐶2 = cos 훼

푂퐶3 = tan 훼

푂퐶4 = cot 훼

푂퐶5 = sec 훼

푂퐶6 = csc 훼

Figure 11-04: A circular angle equal of one circular radian expressed by the area of the circular sector OMA. If OA = 1 then also the area of OMAM’ = 1. Also AM = AM’ and Area(OMA) = Area(OAM’) = 1/2.

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3. Hyperbolic Angles – Hyperbolic Radians

Figure 11-05: Hyperbolic Sector and Hyperbolic Functions

As the point M moves from O on the hyperbola describing the arc AM(going through h1, h2, h3), the x increases from +1 to 푥푀 and y increases from 0 to 푦푀; it describes a hyperbolic sector OAM described by the hyperbolic angle 휃 = 퐴푂푀 . At any position M, the following lines (axis) are orthogonal:

푥 ⊥ 푦

푀퐻1 ⊥ 푦

푀퐻2 ⊥ 푥

퐴퐻3 ⊥ 푥

퐵퐻4 ⊥ 푦

251 At any point during the rotation the following equation holds:

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푦2 − 푥2 = 1 At any position, such as OM, at which the radius makes a circular angle 훽 with the x-axis, the tangent to the hyperbola makes a circular angle of 훽 with the y-axis; or a circular angle of 2훽 with the perpendicular to the radius.

The magnitude of the hyperbolic angle 휃 can be defined in two ways:

1. By the ratio of the hyperbolic arc length s described during the motion by the point M, to the length of the radius 푶푨 = 흆;

Let’s consider again the infinitesimal element of the hyperbolic arc s as being defined by:

푑푠 = √(푑푥)2 + (푑푦)2 and consider the instantaneous value of radius length defined by:

휌 = √푦2 + 푥2 With these notations the element of the angle described during the rotation will be:

푑푠 푑휃 = − ℎ푦푝푒푟푏표푙𝑖푐 푟푎푑𝑖푎푛 휌

This is the infinitesimal element of the hyperbolic angle 휃 shown in Figure 11-05 and it will be expressed in units of hyperbolic radians (later in this section we will define the hyperbolic radian).

As the hyperbolic motion proceeds, from an initial to a final position of the radius OM, the total angle described during the motion is:

푀 푑푠 휃 = ∫ 퐴 휌 If the case of the hyperbola the radius 휌 varies and the above equation becomes:

푀 푑푠 푠 휃 = ∫ = ′ − ℎ푦푝푒푟푏표푙𝑖푐 푟푎푑𝑖푎푛푠 퐴 휌 휌 Where s is the length of the hyperbolic arc AM 휌′ is the integrated mean value of 휌 during the motion, and 휃 is the corresponding hyperbolic

angle expressed in hyperbolic radians (see Figure 11-06).

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Figure 11-06: A Hyperbolic Angle of one hyperbolic radian, in five sections of 0.2 radians each.

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2. By the area of the hyperbolic sector AOM swept out by the radius during the motion from A to M.

Figure 11-07: A hyperbolic angle of one hyperbolic radian expressed by the area of the hyperbolic sector OMA. If OA = 1 then also the area of OMAM’ = 1. Also AM = AM’ and Area(OMA) = Area(OAM’) = 1/2.

It should be stressed here once again that the hyperbolic angle θ of the sector AOM must be distinguished from the circular angle β of the same sector. If the hyperbolic angle θ is equal with one hyperbolic radian, then the corresponding angle β is equal with 0.65 circular radians. Also, the length of an arc of an angle of one hyperbolic radian is equal with 1.3161, if the radius OA is equal with unity.

푅 = √푠𝑖푛ℎ2(1) + 푐표푠ℎ2(1)

푡푎푛ℎ(1) = 0.76159415595576 254

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푎푠𝑖푛(푡푎푛ℎ(1)) = 0.86576948323966 -the gudermanian of 1 hyperbolic radian

푎푠𝑖푛( 푡푎푛ℎ(1) ) = 0.65782955948610- circular radians 1.3161

Figure 11-08: A hyperbolic angle of one hyperbolic radian expressed by ′ ′ the area of the hyperbolic sector 푢푀푢푀 푀 퐴푀. If OA = 1 then also the ′ area of the sector 푢푀푢푀′푀 퐴푀 = 1.

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There is also another geometrical representation of the hyperbolic angle and the hyperbolic radian that we will discussed in the next chapter. For now Figure 11-8 shows a hyperbolic angle of one radian for a hyperbola whose equation is given by:

1 푢푣 = 2

There is a way to define and to derive all hyperbolic functions using basic geometry. We start from a unit circle (푂퐴 = 1), and from point A we have to construct a unit rectangular hyperbola.

Figure 11-09: Graphic representation of Hyperbolic Functions (Advanced Mathematics for Engineers, by Reddick and Millerii - page 89)

The equations of the circle and the hyperbola are:

푐𝑖푟푐푙푒: 푥2 + 푦2 = 1

2 2 ℎ푦푝푒푟푏표푙푎: 푥 − 푦 = 1 Any point (x, y) on the hyperbola is defined as:

256 푥 = 푂퐻2 푦 = 퐻2푁 = 푂퐻1

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Now, we take H2 as the center of a circle and draw a circular arc from N, with the radius H2N. This circle will intersect the initial unit circle at point M. Next, we draw the line H2M which is tangent to the circle in point M and intersect the y-axis in D. So the angle OMH2 is a right angle. Also we have H2N = H2M, so by definition we have:

풔풊풏풉 풖 = 푯ퟐ푵 = 푯ퟐ푴 = 푶푯ퟏ = 풚

Now taking the right triangle OMH2 we have: 푶푴 = ퟏ 푯ퟐ푴 = 퐬퐢퐧퐡 풖 ퟐ ퟐ ퟏ + (퐬퐢퐧퐡 풖) = 푶푯ퟐ From the fact that: ퟏ + 풔풊풏풉ퟐ 풖 = 풄풐풔풉ퟐ 풖 and also that H2M = x we have: ퟐ ퟐ ퟐ (푶푯ퟐ) = 풄풐풔풉 풖 = 풙 and so: 풄풐풔풉 풖 = 푶푯ퟐ = 풙

Next from: 푯 푵 풔풊풏풉 풖 ퟐ = = 풕풂풏풉 풖 푶푯ퟐ 풄풐풔풉 풖

Taking the similar triangle OH2N and OAH3 we have: 푯 푵 푨푯 ퟐ = ퟑ = 풕풂풏풉 풖 푶푯ퟐ 푶푨 and since OA = 1 this gives: 풕풂풏풉 풖 = 푨푯ퟑ

Next, from the similar triangle OMC2 and OMH2 we have:

푶푪 푶푴 ퟐ = 푶푴 푶푯ퟐ

푶푴 = ퟏ

ퟏ ퟏ 푶푪ퟐ = = 푶푯ퟐ 퐜퐨퐬퐡 풖 and so: 풔풆풄풉 풖 = 푶푪ퟐ

From the similar triangles OMD and OMH2 we have:

푫푴 푶푴 =

푶푴 푯ퟐ푴

푶푴 = ퟏ 57

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ퟏ ퟏ 푫푴 = = 푯ퟐ푴 풔풊풏풉 풖

풄풔풄풉 풖 = 푫푴

Now, from the same similar triangles OMD and OMH2 we have:

푶푫 푶푯 = ퟐ 푶푴 푯ퟐ푴 푶푴 = ퟏ

푶푯 퐜퐨퐬퐡 풖 푶푫 = ퟐ = 푯ퟐ푴 퐬퐢퐧퐡 풖

풄풐풕풉 풖 = 푶푫

This is quite a remarkable geometrical interpretation of the hyperbolic functions. It makes use of the intrinsic connection between the circle and the hyperbola, and also introduces the gudermannian angle that we will discuss in more detail next. For now we just want to mention that the circular angle AOM is called the gudermannian of the hyperbolic angle u and is refers as: 품풅 풖.

This relationship can also be inversed, by saying that the hyperbolic angle u is the antigudermannian of the circular angle AOM.

i Arthur Edwin Kennelly "The application of hyperbolic functions to electrical engineering problems", Elibron Classics series, ISBN 0-543- 96403-5, 2006 ii Reddick and Mille “Advanced Mathematics for Engineers”

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