# Hyperbolic Cosines and Sines Theorems for the Triangle Formed by Arcs of Intersecting Semicircles on Euclidean Plane

Authors: Robert M. Yamaleev
Journal: Journal of Mathematics
 (2013)

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## Abstract

The hyperbolic cosines and sines theorems for the curvilinear triangle bounded by circular arcs of three intersecting circles are formulated and proved by using the general complex calculus. The method is based on a key formula establishing a relationship between exponential function and the cross-ratio. The proofs are carried out on Euclidean plane.


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## Body

## 1. Introduction

Classically, the models of the hyperbolic plane are regarded as based on Euclidean geometry. One starts with a piece of a Euclidean plane, a half-plane, or a circular disk and then, in that half-plane or disk the notions of points, lines, distances, angles are defined as things that could be described in terms of Euclidean geometry [1–3]. The model of the hyperbolic plane is the half-plane model. The underlying space of this model is the upper half-plane model H in the complex plane C, defined to be
(1)H={z∈C:Im(z)>0}.
In coordinates (x,y) the line element is defined as
(2)ds2=1y2(dx2+dy2).
The geodesics of this space are semicircles centered on the x-axis and vertical half-lines. The geometrical properties of the figures on the half-plane are studied by considering quantities invariant under an action of the general Möbius group, which consists of compositions of Möbius transformations and reflections [4]. The curvilinear triangle formed by circular arcs of three intersecting semicircles is one of the principal figures of the upper half-plane model H. The hyperbolic laws of sines-cosines for that triangle are proved by using properties of the Möbius group and the upper half-plane H.In this paper we suggest another way of construction of proofs of the sines-cosines theorems of the Poincaré model. The curvilinear triangle formed by circular arcs is the figure of the Euclidean plane; consequently, on the Euclidean plane we have to find relationships antecedent to the sines-cosines hyperbolic laws. Therefore, first of all, we establish these relationships by making use of axioms of the Euclidean plane, only. Secondly, we proove that these relationships can be formulated as the hyperbolic sine-cosine theorems. For that purpose we refer to the general complex calculus and within its framework establish a relationship between exponential function and the cross-ratio. In this way the hyperbolic trigonometry emerges on Euclidean plane in a natural way.The paper is organized as follows. In Section2  a key formula connecting the hyperbolic calculus with the cross-ratio is established. By employing the key formula and the Pythagoras theorem the elements of the right-angled triangle are expressed as functions of the hyperbolic trigonometry. In Section 3 we explore Euclidian properties of the curvilinear triangle bounded by circular arcs of three intersecting semicircles. Two main relationships connecting three intersecting semicircles are established. In Section 4, we prove the theorem of cosines for the curvilinear triangle bounded by the circular arcs. In Section 5, the hyperbolic law of sines and second hyperbolic law of cosines are derived on the basis of the main relationships between these semicircles.


## 2. Hyperbolic Trigonometry in Euclidean Geometry

### 2.1. Trigonometry Induced by General Complex Algebra

The simplest generalization of the complex algebra, denominated asGeneral Complex Algebra, is defined by unique generator e satisfying the quadratic equation [5]
(3)e2-b1e+b0=0,
where coefficients b0,b1 are given by real positive numbers and b12-4b0>0.The following Euler formula holds true:(4)exp(eϕ)=g0(ϕ)+eg1(ϕ).
Denote by x1,x2∈R roots of the quadratic equation (3). The Euler formula (4) is decomposed into two independent equations:
(5)exp(xkϕ)=g0(ϕ)+xkg1(ϕ),k=1,2,
from which one may find explicit formulae for g-functions. These functions are linear combinations of the exponential functions exp(xkϕ), k=1,2. Geometrical interpretation of the general complex algebra is done in [6].Form the following ratio(6)exp((x2-x1)ϕ)=x2-D(ϕ)x1-D(ϕ),
where
(7)D(ϕ)=-g0(ϕ)g1(ϕ).
Introduce a pair of variables X1,X2 by
(8)Xk(ϕ)=xk-D(ϕ),k=1,2.
Then, (6) is written as follows:
(9)exp((X2-X1)ϕ)=X2(ϕ)X1(ϕ).
This formula we denominate as the key formula and use it in order to introduce hyperbolic trigonometry on Euclidean plane. Notice that the argument of the exponential function is proportional to the difference between the numerator and the denominator, X1-X2=x1-x2, and this difference does not depend on the parameter ϕ.


### 2.2. Elements of Right-Angled Triangle as Functions of a Hyperbolic Trigonometry

Let▵ABC be a right-angled triangle with right angle at C. Denote the sides by a,b, the hypotenuse by c, and the angles opposite to a, b, c by A, B, C, correspondingly. Traditionally interrelations between angles and sides of a triangle are described by the trigonometry via periodical sine-cosine functions. The periodical functions of the circular angles are defined via the ratios
(10)sinB=bc,tanB=ba.
Notice that these two ratios are functions of the same angle B. On making use of formula (9) we are able to introduce the hyperbolic trigonometry besides the circular trigonometry. Define the following relationships:
(11)c+ac-a=exp(2ξ).
From this equation it follows that
(12)ac=tanh(ξ),ab=sinh(ξ).
In this way we arrive to the following interrelations between circular and hyperbolic functions:
(13)cosB=ac=tanh(ξ),tanB=ba=1sinh(ξ).Now let us introduce the following geometrical motion; namely, change position of the pointA along the line AC (see, Figure 1). Then, the sides b and c will change while the side a will not change. Since the length a is a constant of this evolution, in agreement with key formula (9) we rewrite (11) as follows:
(14)c+ac-a=exp(2aϕ),
that is, the argument of exponential function is proportional to a: ξ=aϕ, where ϕ is a parameter of the evolution. Formulae in (13) are rewritten as
(15)cosB=ac=tanh(aϕ),tanB=ba=1sinh(aϕ).
From these formulas it is derived that
(16)cotB=sinh(aϕ),
or,
(17)tanB2=exp(-aϕ).Figure 1
Motion of the hypothenuse of the right angled triangle.Install the triangle▵ABC in such a way that the side b=AC lies on the line L, and the side a=BC is perpendicular to this line at the point CIn Figure1 the line AB moves in such a way that cuts line L at points A, A′, A′′, and so forth. Now, let us recall the problem of parallel lines in geometry [7]. Let the ray AB tend to a definite limiting position; in Figure 1 this position is given by line L1. As the point A moves along line L away from point C there are two possibilities to consider.(1)
In Euclidean geometry, the angle between linesL1 and BC is equal to right angle.(2)
The hypothesis of hyperbolic geometry is that this angle is less than the right angle.The most fundamental formula of the hyperbolic geometry is the formula connecting the angle of parallelismΠ(a) and the length a of the perpendicular from the given point to the given line. In order to establish those relationships the concept of horocycles, some circles with center and axis at infinity, was introduced [8]. The great theorem which enables one to introduce the circular functions, sines, and cosines of an angle is that the geometry of shortest lines (horocycles) traced on horosphere is the same as plane Euclidean geometry. The function connecting the angle of parallelism with the distance a is given by
(18)exp(-aκ)=tanΠ(a)2.
Now, we introduce the value inverse to ϕ by K=1/ϕ and rewrite (17) in the form
(19)exp(-aK)=tanB2.Let the pointA run away from the point C to infinity. The following two cases can be considered.(1)
ϕ tends to zero, and  K tends to infinity; then the angle B will tend to right angle.This is true in Euclidean plane.(2)
Suppose thatϕ, and K and the angle B go to some limited values,
(20)limAC⟶∞K=κ,limB=Π(a).In this way, we have established connection of formula (17) with the main formula of hyperbolic geometry (19).


### 2.3. Rotational Motion of a Line Tangent to the Semicircle

The concept of the circular angle in Euclidean plane is intimately related to the figure of a circle and to motion of a point along the circumference. The hyperbolic angle is also related to the circle because of a motion along the circumference coherence with the motion along the hyperbola [9].Consider semicircleC (Figure 2) with end-points and the center on x-axis. Denote by B the center and by K1, K2 the end-points of the semicircle. Through end-points of the semicircle, K1, K2 erect the lines parallel to vertical axis, Y-axis. Draw a line tangent to the semicircle at the point C; this line crosses x-axis at the point A and intersects with the vertical lines at points P1 and P2. Draw a line parallel to Y-axis from the center B which crosses C at the top point N.Figure 2
Semicircle and lines tangent to the semicircle.Denote byr radius of the circle, so that r=BN=BC and r=K1K2. Denote by B the angle ∠ABC. The triangle ▵ABC is a right triangle, so that
(21)(AB)2-(AC)2=r2.
Consider rotational motion of the line tangent to the semicircle at the point C. In Figure 2, two positions of this line are given by lines A′C2 and AC1. When the point Cruns from end point K1 to the top-point N, the point A runs along x-axis from point K1 to infinity. During this motion the triangle formed by the line tangent to the semicircle, the x-axis and the radius of the circle remains to be right angled. This is exactly the case considered in Section 2.2, consequently, we can apply formula (13). According to (13) we write
(22)AB=rcoth(rϕ)=rcosB,AC1=rsinh(rϕ)=rtanB.
Since ▵ABC1~▵AK1P1, we have
(23)P1K1AK1=rAC1=sinh(rϕ),(24)P1K1=AK1sinh(rϕ)=(AB-r)sinh(rϕ)=rexp(-rϕ),P2K2=rexp(rϕ).
On the basis of obtained formulae the following relationships between circular and hyperbolic trigonometric functions are established:
(25)sin(B)=1cosh(rϕ),cot(B)=sinh(rϕ),cos(B)=tanh(rϕ).
If we make the point C tend to the top of semicircle N, the hyperbolic angle ϕ will tend to zero. When the point C tends to end-point K1, the hyperbolic angle tends to infinity. Thus, the hyperbolic angle is measured from the point N, the top of the semicircle. Consider two different positions of the tangent line, corresponding to two positions C1, C2, with hyperbolic angles ϕ2 and ϕ1. According to key formula (6) we write
(26)exp((x2-x1)ϕ2)=x2-D2x1-D2,exp((x2-x1)ϕ1)=x2-D1x1-D1.
The hyperbolic angle ϕ2 corresponds to the arc ⌢NC2, and the hyperbolic angle ϕ1 corresponds to the arc ⌢NC1. Then we suppose that the difference in the hyperbolic angles ϕ2-ϕ1 will correspond to the arc ⌢C2C1. From (26) it follows that
(27)exp(x2-x1)(ϕ2-ϕ1)=x2-D2x1-D2x1-D1x2-D1,Dk=D(ϕk),k=1,2.
It is seen, in the right-hand side we have the cross-ratio. Now, let recall definition of the distance between two points of the geodesic line in the Poincaré model. Let z,w∈H, and let z1 and w1 be end points of a geodesic line passing through z and w. Then the distance between these points is defined by formula
(28)ρ(z1,z3)=log(z1,z2;z3,z4),
where
(29)(z1,z2;z3,z4):=z2-z1z4-z1z4-z3z2-z3
is the cross-ratio. Comparing (29) with (27) we come to the following correspondence: z2=x2, z4=x1, z1=D2, z3=D1. Notice, however, in our construction Dk, k=1,2 are projections of z1, z3 on X-axis. The semicircle C is the geodesic line, and x1, x2 are end-points of the geodesic line.


## 2.1. Trigonometry Induced by General Complex Algebra

The simplest generalization of the complex algebra, denominated asGeneral Complex Algebra, is defined by unique generator e satisfying the quadratic equation [5]
(3)e2-b1e+b0=0,
where coefficients b0,b1 are given by real positive numbers and b12-4b0>0.The following Euler formula holds true:(4)exp(eϕ)=g0(ϕ)+eg1(ϕ).
Denote by x1,x2∈R roots of the quadratic equation (3). The Euler formula (4) is decomposed into two independent equations:
(5)exp(xkϕ)=g0(ϕ)+xkg1(ϕ),k=1,2,
from which one may find explicit formulae for g-functions. These functions are linear combinations of the exponential functions exp(xkϕ), k=1,2. Geometrical interpretation of the general complex algebra is done in [6].Form the following ratio(6)exp((x2-x1)ϕ)=x2-D(ϕ)x1-D(ϕ),
where
(7)D(ϕ)=-g0(ϕ)g1(ϕ).
Introduce a pair of variables X1,X2 by
(8)Xk(ϕ)=xk-D(ϕ),k=1,2.
Then, (6) is written as follows:
(9)exp((X2-X1)ϕ)=X2(ϕ)X1(ϕ).
This formula we denominate as the key formula and use it in order to introduce hyperbolic trigonometry on Euclidean plane. Notice that the argument of the exponential function is proportional to the difference between the numerator and the denominator, X1-X2=x1-x2, and this difference does not depend on the parameter ϕ.


## 2.2. Elements of Right-Angled Triangle as Functions of a Hyperbolic Trigonometry

Let▵ABC be a right-angled triangle with right angle at C. Denote the sides by a,b, the hypotenuse by c, and the angles opposite to a, b, c by A, B, C, correspondingly. Traditionally interrelations between angles and sides of a triangle are described by the trigonometry via periodical sine-cosine functions. The periodical functions of the circular angles are defined via the ratios
(10)sinB=bc,tanB=ba.
Notice that these two ratios are functions of the same angle B. On making use of formula (9) we are able to introduce the hyperbolic trigonometry besides the circular trigonometry. Define the following relationships:
(11)c+ac-a=exp(2ξ).
From this equation it follows that
(12)ac=tanh(ξ),ab=sinh(ξ).
In this way we arrive to the following interrelations between circular and hyperbolic functions:
(13)cosB=ac=tanh(ξ),tanB=ba=1sinh(ξ).Now let us introduce the following geometrical motion; namely, change position of the pointA along the line AC (see, Figure 1). Then, the sides b and c will change while the side a will not change. Since the length a is a constant of this evolution, in agreement with key formula (9) we rewrite (11) as follows:
(14)c+ac-a=exp(2aϕ),
that is, the argument of exponential function is proportional to a: ξ=aϕ, where ϕ is a parameter of the evolution. Formulae in (13) are rewritten as
(15)cosB=ac=tanh(aϕ),tanB=ba=1sinh(aϕ).
From these formulas it is derived that
(16)cotB=sinh(aϕ),
or,
(17)tanB2=exp(-aϕ).Figure 1
Motion of the hypothenuse of the right angled triangle.Install the triangle▵ABC in such a way that the side b=AC lies on the line L, and the side a=BC is perpendicular to this line at the point CIn Figure1 the line AB moves in such a way that cuts line L at points A, A′, A′′, and so forth. Now, let us recall the problem of parallel lines in geometry [7]. Let the ray AB tend to a definite limiting position; in Figure 1 this position is given by line L1. As the point A moves along line L away from point C there are two possibilities to consider.(1)
In Euclidean geometry, the angle between linesL1 and BC is equal to right angle.(2)
The hypothesis of hyperbolic geometry is that this angle is less than the right angle.The most fundamental formula of the hyperbolic geometry is the formula connecting the angle of parallelismΠ(a) and the length a of the perpendicular from the given point to the given line. In order to establish those relationships the concept of horocycles, some circles with center and axis at infinity, was introduced [8]. The great theorem which enables one to introduce the circular functions, sines, and cosines of an angle is that the geometry of shortest lines (horocycles) traced on horosphere is the same as plane Euclidean geometry. The function connecting the angle of parallelism with the distance a is given by
(18)exp(-aκ)=tanΠ(a)2.
Now, we introduce the value inverse to ϕ by K=1/ϕ and rewrite (17) in the form
(19)exp(-aK)=tanB2.Let the pointA run away from the point C to infinity. The following two cases can be considered.(1)
ϕ tends to zero, and  K tends to infinity; then the angle B will tend to right angle.This is true in Euclidean plane.(2)
Suppose thatϕ, and K and the angle B go to some limited values,
(20)limAC⟶∞K=κ,limB=Π(a).In this way, we have established connection of formula (17) with the main formula of hyperbolic geometry (19).


## 2.3. Rotational Motion of a Line Tangent to the Semicircle

The concept of the circular angle in Euclidean plane is intimately related to the figure of a circle and to motion of a point along the circumference. The hyperbolic angle is also related to the circle because of a motion along the circumference coherence with the motion along the hyperbola [9].Consider semicircleC (Figure 2) with end-points and the center on x-axis. Denote by B the center and by K1, K2 the end-points of the semicircle. Through end-points of the semicircle, K1, K2 erect the lines parallel to vertical axis, Y-axis. Draw a line tangent to the semicircle at the point C; this line crosses x-axis at the point A and intersects with the vertical lines at points P1 and P2. Draw a line parallel to Y-axis from the center B which crosses C at the top point N.Figure 2
Semicircle and lines tangent to the semicircle.Denote byr radius of the circle, so that r=BN=BC and r=K1K2. Denote by B the angle ∠ABC. The triangle ▵ABC is a right triangle, so that
(21)(AB)2-(AC)2=r2.
Consider rotational motion of the line tangent to the semicircle at the point C. In Figure 2, two positions of this line are given by lines A′C2 and AC1. When the point Cruns from end point K1 to the top-point N, the point A runs along x-axis from point K1 to infinity. During this motion the triangle formed by the line tangent to the semicircle, the x-axis and the radius of the circle remains to be right angled. This is exactly the case considered in Section 2.2, consequently, we can apply formula (13). According to (13) we write
(22)AB=rcoth(rϕ)=rcosB,AC1=rsinh(rϕ)=rtanB.
Since ▵ABC1~▵AK1P1, we have
(23)P1K1AK1=rAC1=sinh(rϕ),(24)P1K1=AK1sinh(rϕ)=(AB-r)sinh(rϕ)=rexp(-rϕ),P2K2=rexp(rϕ).
On the basis of obtained formulae the following relationships between circular and hyperbolic trigonometric functions are established:
(25)sin(B)=1cosh(rϕ),cot(B)=sinh(rϕ),cos(B)=tanh(rϕ).
If we make the point C tend to the top of semicircle N, the hyperbolic angle ϕ will tend to zero. When the point C tends to end-point K1, the hyperbolic angle tends to infinity. Thus, the hyperbolic angle is measured from the point N, the top of the semicircle. Consider two different positions of the tangent line, corresponding to two positions C1, C2, with hyperbolic angles ϕ2 and ϕ1. According to key formula (6) we write
(26)exp((x2-x1)ϕ2)=x2-D2x1-D2,exp((x2-x1)ϕ1)=x2-D1x1-D1.
The hyperbolic angle ϕ2 corresponds to the arc ⌢NC2, and the hyperbolic angle ϕ1 corresponds to the arc ⌢NC1. Then we suppose that the difference in the hyperbolic angles ϕ2-ϕ1 will correspond to the arc ⌢C2C1. From (26) it follows that
(27)exp(x2-x1)(ϕ2-ϕ1)=x2-D2x1-D2x1-D1x2-D1,Dk=D(ϕk),k=1,2.
It is seen, in the right-hand side we have the cross-ratio. Now, let recall definition of the distance between two points of the geodesic line in the Poincaré model. Let z,w∈H, and let z1 and w1 be end points of a geodesic line passing through z and w. Then the distance between these points is defined by formula
(28)ρ(z1,z3)=log(z1,z2;z3,z4),
where
(29)(z1,z2;z3,z4):=z2-z1z4-z1z4-z3z2-z3
is the cross-ratio. Comparing (29) with (27) we come to the following correspondence: z2=x2, z4=x1, z1=D2, z3=D1. Notice, however, in our construction Dk, k=1,2 are projections of z1, z3 on X-axis. The semicircle C is the geodesic line, and x1, x2 are end-points of the geodesic line.


## 3. Relationships between Elements of Three Intersecting Semicircles

### 3.1. Hyperbolic Cosine-Sine Functions of Arcs of the Semicircle

The key formula (6) admits to define hyperbolic trigonometric functions of the arcs originated from the top N of the semicircle C. In order to determine trigonometric functions of the arcs with arbitrary end-points on the semicircle we have to use formula (27). Consider arc ⌢C1C2 is defined in the first quadrant of the semicircle with end-points at C1 and C2. The arc ⌢C1C2 can be presented as difference of two arcs, both originated from the top of the semicircle:
(30)⌢C1C2=⌢NC1-⌢NC2.
Denote by a1, a2 the angles formed by radiuses BC1 and BC2 with x-axis, correspondingly, where B is a center of the circle. Denote the hyperbolic angle corresponding to the arcs ⌢NCk, k=1,2 by ξ(NCk), k=1,2. Then the functions
(31)coshξ(NC1),coshξ(NC2),sinhξ(NC1),sinhξ(NC2)
are expressed via circular trigonometric functions according to formulae (25):
(32)coshξ(NC1)=1sina1,coshξ(NC2)=1sina2,sinhξ(NC1)=cota1,sinhξ(NC2)=cota2.
Then, by taking into account additional formulae
(33)coshξ(C1C2)=coshξ(NC1)coshξ(NC2)-sinhξ(NC1)sinhξ(NC2),sinhξ(C1C2)=sinhξ(NC1)coshξ(NC2)-coshξ(NC1)sinhξ(NC2),
we get hyperbolic trigonometric functions of the arcs with arbitrary end-points on the semicircle expressed via periodic trigonometric functions:(34a)coshξ(C1C2)=1-cosa1cosa2sina1sina2,sinhξ(C1C2)=cosa1-cosa2sina1sina2.
Compare with analogous formula from Poincaré model ([3], formula (1.2.6)) given by
(34b)coshρ(z,w)=1+|z-w|22Im(z)Im(w).Two complex numbers
(35)z=v0+iv,w=w0+iw
are related to the radius of the semicircle and the angles a1,a2 by
(36)v=rsina2,v0=rcosa2,w=rsina1,w0=rsina1.
Then,
(37)coshξ(C1C2)=1-cosa1cosa2sina1sina2=r2-v0w0wv=1+(v0-w0)2+(v-w)22wv=coshρ(z,w).


### 3.2. Relationships between Elements of the Triangle Bounded by Arcs of the Intersecting Semicircles

In Figure3 three intersecting semicircles with centers installed on horizontal axis at the points Oa, Ob, and Oc are presented. Intersections of the semicircles form triangle ▵~ABC bounded by the arcs ⌢BC,⌢AB,⌢AC. For each arc we can put in correspondence the hyperbolic angle. Denote the hyperbolic angles by a,b,c, where a, b, c consecutively will correspond to the arcs ⌢BC,⌢AC,⌢AB.Figure 3
Triangle formed by arcs of intersecting semicircles.Connect vertices of the triangle with centers of the circle by corresponding radiuses. Denote byak, bk, ck, k=1,2 the angles bounded by the radiuses and x-axis, where a1>a2>0, b1>b2>0, c1>c2>0. By making use of (34a) define hyperbolic cosine-sine functions corresponding to bounding segments:(38a)cosha=1-cosa1cosa2sina1sina2,sinha=cosa1-cosa2sina1sina2,(38b)coshb=1-cosb1cosb2sinb1sinb2,sinhb=cosb1-cosb2sinb1sinb2,(38c)coshc=1-cosc1cosc2sinc1sinc2,sinhc=cosc1-cosc2sinc1sinc2.The usual notion of the angle is used that is, the angle between two curves is defined as an angle between their tangent lines. Let the angles α,β,δ be angles at the verteices A,B,C, correspondingly. For these angles we can define thier proper cosine and sine functions. The angles of the triangle ▵~ABCα,β,δ are closely related to angles a1,a2,b1,b2,c1,c2. From Figure 3 we find the following relationships between them:
(39)β=a2+c2,δ=π-a1-b2,α=b1-c1.
Then,(40a)cosα=cosb1cosc1+sinb1sinc1,(40b)cosβ=cosa2cosc2-sina2sinc2,(40c)cosδ=-cosb2cosa1+sinb2sina1,(41a)sinα=sinb1cosc1-cosb1sinc1,(41b)sinβ=sina2cosc2+cosa2sinc2,(41c)sinδ=sinb2cosa1+cosb2sina1.Denote distances between centers by(42)Ocb=OcOb,Oba=ObOa,Oac=OaOc.
The theorem of sines employed for triangles OcAOb, ObCOa, OcBOa gives six relations of type(43a)sinαOcb=sinc1rb=sinb1rc,(43b)sinδOba=sina1rb=sinb2ra,(43c)sinβOac=sinc2ra=sina2rc.From these relations the first set of main relationships follows.Relation 1.
Consider(44)rasina1=rbsinb2,rcsinc2=rasina2,rcsinc1=rbsinb1.From the draught in Figure 3 it is seen that
(45)Oac=Oba+Ocb,
where
(46)Oac=rccosc2+racosa2,Oba=racosa1+rbcosb2.
Hence,
(47)Ocb=rccosc2+racosa2-racosa1-rbcosb2.
From vertices of ▵~ABC erect lines perpendicular to horizontal line intersect with x-axis at points hA,hB,hC, correspondingly. From Figure 3 we find that
(48)Ocb=OchA-hAOb=rccosc1-rbcosb1.
By equating (47) with (48) we arrive to another main relationship between radii and angles.Relation 2.
One has(49)rccosc1-rbcosb1=rccosc2+racosa2-racosa1-rbcosb2.
We will effect a simplification by using the following designations:(50)racosa1=w01,rasina1=w1,racosa2=v01,rasina2=v1,rbcosb1=w02,rbsinb1=w2,rbcosb2=v02,rbsinb2=v2,rccosc1=w03,rcsinc1=w3,rccosc2=v03,rcsinc2=v3.
In these designations formulae (40a), (40b), (40c) and (41a), (41b), (41c) are written as follows:(51a)rbrcsinα=w2w03-w02w3,rbrccosα=w02w03+w2w3,(51b)rarcsinβ=v1v03+v01v3,rarccosβ=v01v03-v1v3,(51c)rarbsinδ=v2w01+v02w1,rarbcosδ=v2w1-v02w01.Formulae (38a), (38b), (38c) for hyperbolic sines and cosines are rewritten as follows:
(52)cosha=ra2-w01v01w1v1,sinha=raw01-v01w1v1,coshb=rb2-w02v02w2v2,sinhb=rbw02-v02w2v2,coshc=rc2-w03v03w3v3,sinhc=rcw03-v03w3v3.
In the designations equations of Relation 1 are written as
(53)x:=w2=w3,y:=v1=v3,z:=w1=v2.
Equations (43a), (43b), (43c)–(46) are rewritten as follows:
(54)Ocb=w03-w02,Oac=v03+v01,Oba=w01+v02.
Correspondingly, Relation 2 takes the form
(55)w03-w02=v03+v01-w01-v02.
This expresses the fact that Oca is a sum of Ocb and Oba. Notice that (55) can be rewritten also in another equivalent form namely,
(56)w03-v03=w02-v02-(w01-v01).
Denote the segments projections of sides of ▵~ABC on x-axis by P(AC)=hAhC, P(AB)=hAhB, P(BC)=hBhC. From Figure 3 it is seen that
(57)P(AC)=P(AB)+P(BC),
where
(58)P(BC)=w01-v01,P(AC)=w02-v02,P(AB)=w03-v03.


## 3.1. Hyperbolic Cosine-Sine Functions of Arcs of the Semicircle

The key formula (6) admits to define hyperbolic trigonometric functions of the arcs originated from the top N of the semicircle C. In order to determine trigonometric functions of the arcs with arbitrary end-points on the semicircle we have to use formula (27). Consider arc ⌢C1C2 is defined in the first quadrant of the semicircle with end-points at C1 and C2. The arc ⌢C1C2 can be presented as difference of two arcs, both originated from the top of the semicircle:
(30)⌢C1C2=⌢NC1-⌢NC2.
Denote by a1, a2 the angles formed by radiuses BC1 and BC2 with x-axis, correspondingly, where B is a center of the circle. Denote the hyperbolic angle corresponding to the arcs ⌢NCk, k=1,2 by ξ(NCk), k=1,2. Then the functions
(31)coshξ(NC1),coshξ(NC2),sinhξ(NC1),sinhξ(NC2)
are expressed via circular trigonometric functions according to formulae (25):
(32)coshξ(NC1)=1sina1,coshξ(NC2)=1sina2,sinhξ(NC1)=cota1,sinhξ(NC2)=cota2.
Then, by taking into account additional formulae
(33)coshξ(C1C2)=coshξ(NC1)coshξ(NC2)-sinhξ(NC1)sinhξ(NC2),sinhξ(C1C2)=sinhξ(NC1)coshξ(NC2)-coshξ(NC1)sinhξ(NC2),
we get hyperbolic trigonometric functions of the arcs with arbitrary end-points on the semicircle expressed via periodic trigonometric functions:(34a)coshξ(C1C2)=1-cosa1cosa2sina1sina2,sinhξ(C1C2)=cosa1-cosa2sina1sina2.
Compare with analogous formula from Poincaré model ([3], formula (1.2.6)) given by
(34b)coshρ(z,w)=1+|z-w|22Im(z)Im(w).Two complex numbers
(35)z=v0+iv,w=w0+iw
are related to the radius of the semicircle and the angles a1,a2 by
(36)v=rsina2,v0=rcosa2,w=rsina1,w0=rsina1.
Then,
(37)coshξ(C1C2)=1-cosa1cosa2sina1sina2=r2-v0w0wv=1+(v0-w0)2+(v-w)22wv=coshρ(z,w).


## 3.2. Relationships between Elements of the Triangle Bounded by Arcs of the Intersecting Semicircles

In Figure3 three intersecting semicircles with centers installed on horizontal axis at the points Oa, Ob, and Oc are presented. Intersections of the semicircles form triangle ▵~ABC bounded by the arcs ⌢BC,⌢AB,⌢AC. For each arc we can put in correspondence the hyperbolic angle. Denote the hyperbolic angles by a,b,c, where a, b, c consecutively will correspond to the arcs ⌢BC,⌢AC,⌢AB.Figure 3
Triangle formed by arcs of intersecting semicircles.Connect vertices of the triangle with centers of the circle by corresponding radiuses. Denote byak, bk, ck, k=1,2 the angles bounded by the radiuses and x-axis, where a1>a2>0, b1>b2>0, c1>c2>0. By making use of (34a) define hyperbolic cosine-sine functions corresponding to bounding segments:(38a)cosha=1-cosa1cosa2sina1sina2,sinha=cosa1-cosa2sina1sina2,(38b)coshb=1-cosb1cosb2sinb1sinb2,sinhb=cosb1-cosb2sinb1sinb2,(38c)coshc=1-cosc1cosc2sinc1sinc2,sinhc=cosc1-cosc2sinc1sinc2.The usual notion of the angle is used that is, the angle between two curves is defined as an angle between their tangent lines. Let the angles α,β,δ be angles at the verteices A,B,C, correspondingly. For these angles we can define thier proper cosine and sine functions. The angles of the triangle ▵~ABCα,β,δ are closely related to angles a1,a2,b1,b2,c1,c2. From Figure 3 we find the following relationships between them:
(39)β=a2+c2,δ=π-a1-b2,α=b1-c1.
Then,(40a)cosα=cosb1cosc1+sinb1sinc1,(40b)cosβ=cosa2cosc2-sina2sinc2,(40c)cosδ=-cosb2cosa1+sinb2sina1,(41a)sinα=sinb1cosc1-cosb1sinc1,(41b)sinβ=sina2cosc2+cosa2sinc2,(41c)sinδ=sinb2cosa1+cosb2sina1.Denote distances between centers by(42)Ocb=OcOb,Oba=ObOa,Oac=OaOc.
The theorem of sines employed for triangles OcAOb, ObCOa, OcBOa gives six relations of type(43a)sinαOcb=sinc1rb=sinb1rc,(43b)sinδOba=sina1rb=sinb2ra,(43c)sinβOac=sinc2ra=sina2rc.From these relations the first set of main relationships follows.Relation 1.
Consider(44)rasina1=rbsinb2,rcsinc2=rasina2,rcsinc1=rbsinb1.From the draught in Figure 3 it is seen that
(45)Oac=Oba+Ocb,
where
(46)Oac=rccosc2+racosa2,Oba=racosa1+rbcosb2.
Hence,
(47)Ocb=rccosc2+racosa2-racosa1-rbcosb2.
From vertices of ▵~ABC erect lines perpendicular to horizontal line intersect with x-axis at points hA,hB,hC, correspondingly. From Figure 3 we find that
(48)Ocb=OchA-hAOb=rccosc1-rbcosb1.
By equating (47) with (48) we arrive to another main relationship between radii and angles.Relation 2.
One has(49)rccosc1-rbcosb1=rccosc2+racosa2-racosa1-rbcosb2.
We will effect a simplification by using the following designations:(50)racosa1=w01,rasina1=w1,racosa2=v01,rasina2=v1,rbcosb1=w02,rbsinb1=w2,rbcosb2=v02,rbsinb2=v2,rccosc1=w03,rcsinc1=w3,rccosc2=v03,rcsinc2=v3.
In these designations formulae (40a), (40b), (40c) and (41a), (41b), (41c) are written as follows:(51a)rbrcsinα=w2w03-w02w3,rbrccosα=w02w03+w2w3,(51b)rarcsinβ=v1v03+v01v3,rarccosβ=v01v03-v1v3,(51c)rarbsinδ=v2w01+v02w1,rarbcosδ=v2w1-v02w01.Formulae (38a), (38b), (38c) for hyperbolic sines and cosines are rewritten as follows:
(52)cosha=ra2-w01v01w1v1,sinha=raw01-v01w1v1,coshb=rb2-w02v02w2v2,sinhb=rbw02-v02w2v2,coshc=rc2-w03v03w3v3,sinhc=rcw03-v03w3v3.
In the designations equations of Relation 1 are written as
(53)x:=w2=w3,y:=v1=v3,z:=w1=v2.
Equations (43a), (43b), (43c)–(46) are rewritten as follows:
(54)Ocb=w03-w02,Oac=v03+v01,Oba=w01+v02.
Correspondingly, Relation 2 takes the form
(55)w03-w02=v03+v01-w01-v02.
This expresses the fact that Oca is a sum of Ocb and Oba. Notice that (55) can be rewritten also in another equivalent form namely,
(56)w03-v03=w02-v02-(w01-v01).
Denote the segments projections of sides of ▵~ABC on x-axis by P(AC)=hAhC, P(AB)=hAhB, P(BC)=hBhC. From Figure 3 it is seen that
(57)P(AC)=P(AB)+P(BC),
where
(58)P(BC)=w01-v01,P(AC)=w02-v02,P(AB)=w03-v03.


## 4. Hyperbolic Law of Cosines I for the Triangle Formed by Intersection of Three Semicircles

The main aim of this section is to prove the hyperbolic law theorem of cosines I for the triangle▵~ABC formed by arcs of intersecting semicircles with centers installed on x-axis (Figure 3). This law is given by the following set of equations:
(59)coshc=coshacoshb-sinhasinhbcosδ,coshb=coshacoshc-sinhcsinhacosβ,cosha=coshccoshb-sinhcsinhbcosα.Theorem 3 (theorem of cosines I).
The following equation for elements of the triangle▵~ABC formed by arcs of three intersecting semicircles holds true
(60)coshc=coshacoshb-sinhasinhbcosδ,
where coshc,cosha,coshb,sinha,sinhb,cosδ are defined by formulae (51a), (51b), (51c)-(52).Proof.
Square both sides of Relation2 to obtain
(61)(w03-v03)2=(w02-v02)2+(w01-v01)2-2(w02-v02)(w01-v01)
and evaluate this equality by taking into account formulae (51a), (51b), (51c)-(52). First of all, evaluate the left-hand side of this equation as follows:
(62)(w03-v03)2=v032+w032-2v03w03=v032+w032-2rc2+2(rc2-v03w03),
where
(63)2rc2=w032+w32+v032+v32.
Use (63) and write (62) as follows:
(64)v032+w032-2rc2=v032+w032-(w032+w32+v032+v32)=-(w32+v32).
Equation (61) takes the following form:
(65)-(w32+v32)︸+2(rc2-v03w03)=(w02-v02)2+(w01-v01)2-2(w02-v02)(w01-v01).
The underlined term passes to the right-hand side of the equation. Then in the left-hand side remains the expression
(66)2(rc2-v03w03)=2v3w3coshc.
Dividing  both sides of the obtained equation by v3w3, we get
(67)2coshc=1v3w3(w32+v32+(v02-w02)2+(v01-w01)2-2(v02-w02)(v01-w01)(v01-w01)2),
according to Relation 1  w3=w2, v3=v1. These relations make true the following equation
(68)1w3v3=1w2v21w1v1v2w1.
On making use this formula in the right-hand side of (67) we come to the following equation:
(69)2coshc=1w2v21w1v1×{v2w1(w32+v32+(v02-w02)2+(v01-w01)2)}-21w2v21w1v1v2w1(v02-w02)(v01-w01)︸.
Evaluate now the underlined term, which we firstly write as follows:
(70)2(v02-w02)(v01-w01)1w2v21w1v1v2w1=2v2w1rarbrb(v02-w02)w2v2ra(v01-w01)w1v1.
From the second equation of (51c) we have
(71)v2w1rarb=cosδ+v02w01rarb.
By making use of (71) in (70), evaluate (70) as follows:
(72)2v2w1rarbrb(v02-w02)w2v2ra(v01-w01)w1v1=2(cosδ+v02w01rarb)rb(v02-w02)w2v2ra(v01-w01)w1v1=2cosδrb(v02-w02)w2v2ra(v01-w01)w1v1-2v02w01(v02-w02)w2v2(v01-w01)w1v1=2cosδsinhasinhb-2v02w01(v02-w02)w2v2(v01-w01)w1v1.
Replace the underlined term of (69) by (72), and pass expression containing sines and cosines to the left-hand side of the obtained equation. As a result, we come to the following equation:
(73)2coshc-2cosδsinhbsinha=1w2v2w1v1{v2w1(w22+v12+(w02-v02)2+(w01-v01)2)-2v02w01×(w02-v02)(w01-v01)(w01-v01)2}
By applying the elementary algebra one may show that (see, [10])
(74)v2w1(w22+v12+(w02-v02)2+(w01-v01)2)-2v02w01(w02-v02)(w01-v01)=2(ra2-v01w01)(rb2-v02w02).
Thus, in the right-hand side of (73) we have
(75)21w1v1v2w2(ra2-v01w01)(rb2-v02w02),
which according to formulae (52) is nothing else than
(76)2coshacoshb.
By using it in (73) we arrive to the following equation:
(77)2coshc-2coshacoshb=2cosδsinhasinhb.


## 5. Hyperbolic Laws of Sines and Cosines II

The main task of this section is to provehyperbolic law (theorem) of sines, which is given by the formulae
(78)sinhasinα=sinhbsinβ=sinhcsinδ,
and the hyperbolic law (theorem) of cosines II given by the formulae
(79)cosδ=-cosαcosβ-sinαsinβcoshc,(80)cosβ=-cosαcosδ-sinαsinδcoshb,(81)cosα=-cosβcosδ-sinδsinβcosha.


### 5.1. Hyperbolic Theorem of Sines and Its Geometrical Interpretation on Euclidean Plane

Theorem 4.
The ratios of projections of the sides of triangle▵~ABC on x-axis to corresponding distances between centers of the semicircles are equal to each other.Proof.
Projections of the sides of▵~ABC are given by formulae
(82)Pbc=racosa1-racosa2,Pca=rbcosb1-rbcosb2,Pab=rccosc1-rccosc2.
Distances between centers of the circles have been defined as (see, (46) and  (47))
(83)Oca=rccosc2+racosa2,Oba=racosa1+rbcosb2,Ocb=rccosc1-rbcosb1,(84)Pca=Pbc+Pab,Oca=Obc+Oab.
Relation 1 given by the set of equations
(85)rasina1=rbsinb2,rcsinc2=rasina2,rcsinc1=rbsinb1
raises to the second power,
(86)ra2-ra2cos2a1=rb2-rb2cos2b2,rc2-rc2cos2c2=ra2-ra2cos2a2,rc2-rc2cos2c1=rb2-rb2cos2b1.
Then, square distances Oik2,i,k=a,b,c and use (86). We get
(87)Oca2=rc2cos2c2+ra2cos2a2+2rcracosc2cosa2,Oba2=ra2cos2a1+rb2cos2b2+2rarbcosa1cosb2,Ocb2=rc2cos2c1+rb2cos2b1-2rcrbcosc1cosb1.Combine (86) with (87); in this way we come to the following system of equations:(88a)(a)Oca2=rc2-ra2+2racosa2Oac,(b)Oca2=ra2-rc2+2rccosc2Oac,(88b)(a)Oba2=ra2-rb2+2rbcosb2Oba,(b)Oba2=rb2-ra2+2racosa1Oba,(88c)(a)Ocb2=rb2-rc2+2rccosc1Ocb,(b)Ocb2=rc2-rb2-2rbcosb1Ocb.From these equations the cosines of the angles a1,a2,b1,b2,c1,c2 are expressed:
(89)Oba2-rb2+ra22raOba=cosa1,Oca2-rc2+ra22raOac=cosa2,Ocb2-rb2+rc22rcOcb=cosc1,Oca2-ra2+rc22rcOac=cosc2,-Ocb2+rc2-rb22rbOcb=cosb1,Oba2-ra2+rb22rbOba=cosb2.
Having these formulae we may present the projection Pca as follows
(90)Pac=rbcosb1-rbcosb2=-Ocb2+rc2-rb22Ocb-Oba2-ra2+rb22Oba=-Ocb2Oba+rc2Oba-rb2Oba-Oba2Ocb+ra2Ocb-rb2Ocb2OcbOba
The first ratio is presented as follows:
(91)PcaOca=-OcbOba(Ocb+Oba)+rc2Oba-rb2(Oba-Ocb)+ra2Ocb2OcbObaOca=-OcbOba(Oca)+rc2Oba-rb2(Oca)+ra2Ocb2OcbObaOca.
Now, in the same way let us calculate the next ratio, namely, Pbc/Obc. Formula for the projection is evaluated as follows:
(92)Pbc=racosa1-racosa2=Oba2-rb2+ra22Oba-Oca2-rc2+ra22Oac=((ObaOac(Oba-Oca)-rb2Oac+ra2Oac)+rc2Oba-ra2Oba)2OacOba.
By taking into account equation -Ocb=Oba-Oac, we get
(93)Pbc=(ObaOacOcb-rb2Oac+ra2Ocb+rc2Oba)2OacOba.
Next, let us calculate the ratio
(94)PbcObc=(ObaOac(Ocb)-rb2Oac+ra2(Ocb)+rc2Oba)2OacObaObc.
This expression coincides with (91); hence,
(95)PbcObc=PacOac.
By taking into account (84), we arrive to desired relations
(96)PbcObc=PacOac=PabOab.In the sequel come back to designations introduced in Section3. In these designations (96) is written as follows:
(97)w01-v01w03-w02=w02-v02v03+v01=w03-v03w01+v02.Theorem 5.
The sides and the angles of triangle▵~ABC satisfy (78).Proof.
By using formulas (38a), (38b), and (38c) for sinh a, sinh b, sinh c and formulas (41a), (41b), and (41c) for sin α, sinβ,sinδ, we write:
(98)sinhasinα=ra(w01-v01)yz:x(w03-w02)rbrc=w01-v01w03-w02xyzrarbrc,sinhbsinβ=w02-v02rbxz:y(v03+v01)rarc=w02-v02v03+v01xyzrarbrc,sinhcsinδ=w03-v03rcyx:z(w01+v02)rbra=w03-v03w01+v02xyzrarbrc.
It is seen that these equations contain a common factor which is symmetric with respect to a,b,c and x,y,z. Multiply all equations of (97) by this factor. We arrive to (78).Theorem 6.
The sides and the angles of triangle▵~ABC satisfy the following equation:
(99)cosδ=sinαsinβcoshc-cosαcosβ.Proof.
Evaluate the first term of the right-hand side of (99) by making use of (43a), (43b), and (43c):
(100)sinαsinβcoshc=1rarbOcaOcb(1-cosc1cosc2)=1rarbOcaOcb×(1-Ocb2-rb2+rc22rcOcbOca2-ra2+rc22rcOac)=1rarbOacOcb-1rarbOacOcb×(Ocb2-rb2+rc22rcOcbOac2-ra2+rc22rcOac)=14rarbrc2(4OacOcbrc2-Ocb2Oca2-(Ocb2+Oca2)rc2+Ocb2ra2+Oac2rb2).
Now calculate the product cosαcosβ by using the following formulae:
(101)cosα=12rbrc(rc2+rb2-Ocb2),cosβ=-12rarc(rc2+ra2-Oac2).
We obtain
(102)cosαcosβ=12rbrc(rc2+rb2-Ocb2)12rarc(rc2+ra2-Oac2)=14rbrarc2×(Ocb2Oac2-(Ocb2+Oac2)rc2-Ocb2ra2-Oac2rb2).
By using (100) and (102) calculate the difference
(103)sinαsinβcoshc-cosαcosβ=14rarbrc2(4OacOcbrc2-Ocb2Oac2-(Ocb2+Oac2)rc2+Ocb2ra2+Oac2rb2)-14rbrarc2×(+Ocb2Oac2-(Ocb2+Oac2)rc2-Ocb2ra2-Oac2rb2)=12rbra(ra2+rb2-Oab2)=cosδ.
Thus, we got (99).The other two equations, (80) and (81), are proved analogously.


## 5.1. Hyperbolic Theorem of Sines and Its Geometrical Interpretation on Euclidean Plane

Theorem 4.
The ratios of projections of the sides of triangle▵~ABC on x-axis to corresponding distances between centers of the semicircles are equal to each other.Proof.
Projections of the sides of▵~ABC are given by formulae
(82)Pbc=racosa1-racosa2,Pca=rbcosb1-rbcosb2,Pab=rccosc1-rccosc2.
Distances between centers of the circles have been defined as (see, (46) and  (47))
(83)Oca=rccosc2+racosa2,Oba=racosa1+rbcosb2,Ocb=rccosc1-rbcosb1,(84)Pca=Pbc+Pab,Oca=Obc+Oab.
Relation 1 given by the set of equations
(85)rasina1=rbsinb2,rcsinc2=rasina2,rcsinc1=rbsinb1
raises to the second power,
(86)ra2-ra2cos2a1=rb2-rb2cos2b2,rc2-rc2cos2c2=ra2-ra2cos2a2,rc2-rc2cos2c1=rb2-rb2cos2b1.
Then, square distances Oik2,i,k=a,b,c and use (86). We get
(87)Oca2=rc2cos2c2+ra2cos2a2+2rcracosc2cosa2,Oba2=ra2cos2a1+rb2cos2b2+2rarbcosa1cosb2,Ocb2=rc2cos2c1+rb2cos2b1-2rcrbcosc1cosb1.Combine (86) with (87); in this way we come to the following system of equations:(88a)(a)Oca2=rc2-ra2+2racosa2Oac,(b)Oca2=ra2-rc2+2rccosc2Oac,(88b)(a)Oba2=ra2-rb2+2rbcosb2Oba,(b)Oba2=rb2-ra2+2racosa1Oba,(88c)(a)Ocb2=rb2-rc2+2rccosc1Ocb,(b)Ocb2=rc2-rb2-2rbcosb1Ocb.From these equations the cosines of the angles a1,a2,b1,b2,c1,c2 are expressed:
(89)Oba2-rb2+ra22raOba=cosa1,Oca2-rc2+ra22raOac=cosa2,Ocb2-rb2+rc22rcOcb=cosc1,Oca2-ra2+rc22rcOac=cosc2,-Ocb2+rc2-rb22rbOcb=cosb1,Oba2-ra2+rb22rbOba=cosb2.
Having these formulae we may present the projection Pca as follows
(90)Pac=rbcosb1-rbcosb2=-Ocb2+rc2-rb22Ocb-Oba2-ra2+rb22Oba=-Ocb2Oba+rc2Oba-rb2Oba-Oba2Ocb+ra2Ocb-rb2Ocb2OcbOba
The first ratio is presented as follows:
(91)PcaOca=-OcbOba(Ocb+Oba)+rc2Oba-rb2(Oba-Ocb)+ra2Ocb2OcbObaOca=-OcbOba(Oca)+rc2Oba-rb2(Oca)+ra2Ocb2OcbObaOca.
Now, in the same way let us calculate the next ratio, namely, Pbc/Obc. Formula for the projection is evaluated as follows:
(92)Pbc=racosa1-racosa2=Oba2-rb2+ra22Oba-Oca2-rc2+ra22Oac=((ObaOac(Oba-Oca)-rb2Oac+ra2Oac)+rc2Oba-ra2Oba)2OacOba.
By taking into account equation -Ocb=Oba-Oac, we get
(93)Pbc=(ObaOacOcb-rb2Oac+ra2Ocb+rc2Oba)2OacOba.
Next, let us calculate the ratio
(94)PbcObc=(ObaOac(Ocb)-rb2Oac+ra2(Ocb)+rc2Oba)2OacObaObc.
This expression coincides with (91); hence,
(95)PbcObc=PacOac.
By taking into account (84), we arrive to desired relations
(96)PbcObc=PacOac=PabOab.In the sequel come back to designations introduced in Section3. In these designations (96) is written as follows:
(97)w01-v01w03-w02=w02-v02v03+v01=w03-v03w01+v02.Theorem 5.
The sides and the angles of triangle▵~ABC satisfy (78).Proof.
By using formulas (38a), (38b), and (38c) for sinh a, sinh b, sinh c and formulas (41a), (41b), and (41c) for sin α, sinβ,sinδ, we write:
(98)sinhasinα=ra(w01-v01)yz:x(w03-w02)rbrc=w01-v01w03-w02xyzrarbrc,sinhbsinβ=w02-v02rbxz:y(v03+v01)rarc=w02-v02v03+v01xyzrarbrc,sinhcsinδ=w03-v03rcyx:z(w01+v02)rbra=w03-v03w01+v02xyzrarbrc.
It is seen that these equations contain a common factor which is symmetric with respect to a,b,c and x,y,z. Multiply all equations of (97) by this factor. We arrive to (78).Theorem 6.
The sides and the angles of triangle▵~ABC satisfy the following equation:
(99)cosδ=sinαsinβcoshc-cosαcosβ.Proof.
Evaluate the first term of the right-hand side of (99) by making use of (43a), (43b), and (43c):
(100)sinαsinβcoshc=1rarbOcaOcb(1-cosc1cosc2)=1rarbOcaOcb×(1-Ocb2-rb2+rc22rcOcbOca2-ra2+rc22rcOac)=1rarbOacOcb-1rarbOacOcb×(Ocb2-rb2+rc22rcOcbOac2-ra2+rc22rcOac)=14rarbrc2(4OacOcbrc2-Ocb2Oca2-(Ocb2+Oca2)rc2+Ocb2ra2+Oac2rb2).
Now calculate the product cosαcosβ by using the following formulae:
(101)cosα=12rbrc(rc2+rb2-Ocb2),cosβ=-12rarc(rc2+ra2-Oac2).
We obtain
(102)cosαcosβ=12rbrc(rc2+rb2-Ocb2)12rarc(rc2+ra2-Oac2)=14rbrarc2×(Ocb2Oac2-(Ocb2+Oac2)rc2-Ocb2ra2-Oac2rb2).
By using (100) and (102) calculate the difference
(103)sinαsinβcoshc-cosαcosβ=14rarbrc2(4OacOcbrc2-Ocb2Oac2-(Ocb2+Oac2)rc2+Ocb2ra2+Oac2rb2)-14rbrarc2×(+Ocb2Oac2-(Ocb2+Oac2)rc2-Ocb2ra2-Oac2rb2)=12rbra(ra2+rb2-Oab2)=cosδ.
Thus, we got (99).The other two equations, (80) and (81), are proved analogously.


## 6. Concluding Remarks

We have proved three hyperbolic cosine-sine theorems for the triangular formed by arcs of three intersecting semicircles by using only elements of the Euclidean geometry. This geometrical figure is one of the basic figures of the Poincaré model of hyperbolic geometry. Conventionally, in the textbooks on the Poincaré model of hyperbolic geometry these theorems are proved by employing invariance properties of the cross-ratio with respect to Möbius group transformations. In that case one remains with the impression that the hyperbolic cosine-sine theorems are consequences of a special structure of theH half-plane. We have shown that the hyperbolic trigonometry as well as periodic trigonometry arises on Euclidean plane in a natural way.The method of parametrization of the mass-shell equation via the hyperbolic angle,the rapidity, is widely used in the relativistic physics [4]. The quadratic equation (3) is closely related to the mass-shell equation
(104)p02-p2=m2c2,
where mc plays the role of radius of the semicircle 2mc=x2-x1. Within the framework of the present method we come to the following parametrization of energy momentum of the relativistic particle [11]:
(105)p0=mccoth(mcϕ),p=mcsinh(mcϕ),0≤ϕ<∞.
In this way, the method developed in this paper may be used in order to give a new geometrical interpretation of rapidity widely used in relativistic physics (see, e.g., [12] and references therein).



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