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

Hindawi Publishing Corporation Journal of Mathematics Volume 2013, Article ID 920528, 10 pages http://dx.doi.org/10.1155/2013/920528

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

Robert M. Yamaleev1,2 1 Joint Institute for Nuclear Research, Laboratory of Information Technology, Street Joliot Curie 6, CP 141980, P.O. Box 141980, Dubna, Russia 2 Universidad Nacional Autonoma de Mexico, Facultad de Estudios Superiores, Avenida 1 Mayo, 54740 Cuautitlan´ Izcalli, MEX, Mexico

Correspondence should be addressed to Robert M. Yamaleev; [email protected]

Received 17 December 2012; Accepted 21 February 2013

Academic Editor: Sujit Samanta

Copyright © 2013 Robert M. Yamaleev. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

1. Introduction laws of sines-cosines for that triangle are proved by using properties of the Mobius¨ group and the upper half-plane 𝐻. Classically, the models of the hyperbolic plane are regarded In this paper we suggest another way of construction of as based on Euclidean geometry. One starts with a piece of a proofs of the sines-cosines theorems of the Poincaremodel.´ Euclidean plane, a half-plane, or a circular disk and then, in The curvilinear triangle formed by circular arcs is the figure that half-plane or disk the notions of points, lines, distances, of the Euclidean plane; consequently, on the Euclidean plane angles are defined as things that could be described in terms we have to find relationships antecedent to the sines-cosines of Euclidean geometry [1–3]. The model of the hyperbolic plane is the half-plane model. The underlying space of this hyperbolic laws. Therefore, first of all, we establish these model is the upper half-plane model 𝐻 in the complex plane relationships by making use of axioms of the Euclidean plane, 𝐶,definedtobe only. Secondly, we proove that these relationships can be formulated as the hyperbolic sine-cosine theorems. For that 𝐻={𝑧∈𝐶:Im (𝑧) >0} . (1) purpose we refer to the general complex calculus and within its framework establish a relationship between exponential (𝑥, 𝑦) In coordinates thelineelementisdefinedas function and the cross-ratio. In this way the hyperbolic 1 trigonometry emerges on Euclidean plane in a natural way. 𝑑𝑠2 = (𝑑𝑥2 +𝑑𝑦2). 𝑦2 (2) The paper is organized as follows. In Section 2 akey formula connecting the hyperbolic calculus with the cross- The geodesics of this space are semicircles centered on the 𝑥- ratio is established. By employing the key formula and the axis and vertical half-lines. The geometrical properties of the Pythagoras theorem the elements of the right-angled triangle figures on the half-plane are studied by considering quantities are expressed as functions of the hyperbolic trigonometry. In invariant under an action of the general Mobius¨ group,which Section 3 we explore Euclidian properties of the curvilinear consists of compositions of Mobius¨ transformations and triangle bounded by circular arcs of three intersecting semi- reflections4 [ ]. The curvilinear triangle formed by circular circles. Two main relationships connecting three intersecting arcs of three intersecting semicircles is one of the principal semicircles are established. In Section 4,weprovethetheo- figures of the upper half-plane model 𝐻. The hyperbolic rem of cosines for the curvilinear triangle bounded by the 2 Journal of Mathematics

𝐵 𝐿1 This formula we denominate as the key formula and use it 𝐿2 in order to introduce hyperbolic trigonometry on Euclidean 𝐵 II(a) plane. Notice that the argument of the exponential function 𝐵󳰀 is proportional to the difference between the numerator and 󳰀󳰀 𝐵 the denominator, 𝑋1 −𝑋2 =𝑥1 −𝑥2, and this difference does 󳰀󳰀 𝐴 not depend on the parameter 𝜙. 𝐿 𝐴󳰀 𝐴 𝐶 2.2. Elements of Right-Angled Triangle as Functions of a Hyper- Figure 1: Motion of the hypothenuse of the right angled triangle. bolic Trigonometry. Let 󳵻𝐴𝐵𝐶 bearight-angledtrianglewith right angle at 𝐶. Denote the sides by 𝑎,,thehypotenuseby 𝑏 𝑐, and the angles opposite to 𝑎, 𝑏, 𝑐 by 𝐴, 𝐵, 𝐶, correspondingly. circular arcs. In Section 5, the hyperbolic law of sines and Traditionally interrelations between angles and sides of a second hyperbolic law of cosines are derived on the basis of triangle are described by the trigonometry via periodical the main relationships between these semicircles. sine-cosine functions. The periodical functions of the circular angles are defined via the ratios 2. Hyperbolic Trigonometry in 𝑏 𝑏 sin 𝐵= , tan 𝐵= . (10) Euclidean Geometry 𝑐 𝑎 2.1. Trigonometry Induced by General Complex Algebra. The Notice that these two ratios are functions of the same angle 𝐵 simplest generalization of the complex algebra, denominated .Onmakinguseofformula(9)weareabletointroducethe as General Complex Algebra, is defined by unique generator e hyperbolic trigonometry besides the circular trigonometry. satisfying the quadratic equation [5] Define the following relationships: 𝑐+𝑎 e2 −𝑏e +𝑏 =0, = (2𝜉) . 1 0 (3) 𝑐−𝑎 exp (11) where coefficients 𝑏0,𝑏1 are given by real positive numbers 2 From this equation it follows that and 𝑏 −4𝑏0 >0. 1 𝑎 𝑎 The following Euler formula holds true: = (𝜉) , = (𝜉) . 𝑐 tanh 𝑏 sinh (12) exp (e𝜙) =𝑔0 (𝜙) + e𝑔1 (𝜙) . (4) In this way we arrive to the following interrelations between

Denote by 𝑥1,𝑥2 ∈𝑅rootsofthequadraticequation(3). circular and hyperbolic functions: The Euler formula (4) is decomposed into two independent 𝑎 𝑏 1 equations: cos 𝐵= = tanh (𝜉) , tan 𝐵= = . (13) 𝑐 𝑎 sinh (𝜉) exp (𝑥𝑘𝜙) =𝑔0 (𝜙) +𝑥𝑘𝑔1 (𝜙) , 𝑘=1,2, (5) Now let us introduce the following geometrical motion; 𝑔 namely, change position of the point 𝐴 along the line 𝐴𝐶 from which one may find explicit formulae for -functions. 𝑏 𝑐 These functions are linear combinations of the exponential (see, Figure 1). Then, the sides and will change while the side 𝑎 will not change. Since the length 𝑎 is a constant of this functions exp(𝑥𝑘𝜙), 𝑘=1,2. Geometrical interpretation of the general complex algebra is done in [6]. evolution, in agreement with key formula (9)werewrite(11) Form the following ratio as follows: 𝑐+𝑎 = exp (2𝑎𝜙) , (14) 𝑥2 −𝐷(𝜙) 𝑐−𝑎 exp ((𝑥2 −𝑥1)𝜙)= , (6) 𝑥1 −𝐷(𝜙) that is, the argument of exponential function is proportional 𝑎 𝜉=𝑎𝜙 𝜙 where to : ,where is a parameter of the evolution. Formulae in (13) are rewritten as 𝑔0 (𝜙) 𝐷(𝜙)=− . (7) 𝑎 𝑏 1 𝑔1 (𝜙) cos 𝐵= = tanh (𝑎𝜙) , tan 𝐵= = . (15) 𝑐 𝑎 sinh (𝑎𝜙) Introduce a pair of variables 𝑋1, 𝑋2 by From these formulas it is derived that 𝑋𝑘 (𝜙) =𝑘 𝑥 −𝐷(𝜙), 𝑘=1,2. (8) cot 𝐵=sinh (𝑎𝜙) , (16) Then,6 ( ) is written as follows: or, 𝑋 (𝜙) ((𝑋 −𝑋 )𝜙)= 2 . 𝐵 exp 2 1 (9) tan = exp (−𝑎𝜙) . (17) 𝑋1 (𝜙) 2 Journal of Mathematics 3

𝑃2

𝑁 𝐶1

𝑃1 𝐶2

󳰀 𝐴 𝐴 𝐾1 𝐵 𝐾2

Figure 2: Semicircle and lines tangent to the semicircle.

Install the triangle 󳵻𝐴𝐵𝐶 in such a way that the side 𝑏= In this way, we have established connection of formula (17) 𝐴𝐶 lies on the line 𝐿,andtheside𝑎=𝐵𝐶is perpendicular to with the main formula of hyperbolic geometry (19). this line at the point 𝐶 In Figure 1 the line 𝐴𝐵 movesinsuchawaythatcutsline 󸀠 󸀠󸀠 2.3. Rotational Motion of a Line Tangent to the Semicircle. The 𝐿 at points 𝐴, 𝐴 , 𝐴 , and so forth. Now, let us recall the concept of the circular angle in Euclidean plane is intimately problemofparallellinesingeometry[7]. Let the ray 𝐴𝐵 tend relatedtothefigureofacircleandtomotionofapointalong to a definite limiting position; in Figure 1 this position is given the circumference. The hyperbolic angle is also related to the 𝐿 𝐴 𝐿 by line 1.Asthepoint moves along line away from point circle because of a motion along the circumference coherence 𝐶 there are two possibilities to consider. with the motion along the hyperbola [9]. C (1) In Euclidean geometry, the angle between lines 𝐿1 and Consider semicircle (Figure 2) with end-points and 𝑥 𝐵 𝐾 𝐾 𝐵𝐶 is equal to right angle. the center on -axis. Denote by the center and by 1, 2 the end-points of the semicircle. Through end-points of the (2) The hypothesis of hyperbolic geometry is that this semicircle, 𝐾1, 𝐾2 erect the lines parallel to vertical axis, 𝑌- angleislessthantherightangle. axis. Draw a line tangent to the semicircle at the point 𝐶; The most fundamental formula of the hyperbolic geom- this line crosses 𝑥-axis at the point 𝐴 and intersects with the Π(𝑎) etry is the formula connecting the angle of parallelism vertical lines at points 𝑃1 and 𝑃2. Draw a line parallel to 𝑌-axis and the length 𝑎 of the perpendicular from the given point from the center 𝐵 which crosses C at the top point 𝑁. to the given line. In order to establish those relationships Denote by 𝑟 radius of the circle, so that 𝑟=𝐵𝑁=𝐵𝐶and the concept of horocycles, some circles with center and axis 𝑟=𝐾1𝐾2.Denoteby𝐵 the angle ∠𝐴𝐵𝐶.Thetriangle󳵻𝐴𝐵𝐶 at infinity, was introduced [8]. The great theorem which is a right triangle, so that enables one to introduce the circular functions, sines, and (𝐴𝐵)2 − (𝐴𝐶)2 =𝑟2. cosinesofanangleisthatthegeometryofshortestlines (21) (horocycles) traced on horosphere is the same as plane Consider rotational motion of the line tangent to the semi- Euclidean geometry. The function connecting the angle of circle at the point C.InFigure 2,twopositionsofthislineare 𝑎 󸀠 parallelism with the distance is given by given by lines 𝐴 𝐶2 and 𝐴𝐶1. When the point 𝐶runs from 𝑎 Π (𝑎) end point 𝐾1 to the top-point 𝑁,thepoint𝐴 runs along 𝑥- (− )= . (18) 𝐾 exp 𝜅 tan 2 axis from point 1 to infinity. During this motion the triangle 𝑥 𝜙 K =1/𝜙 formed by the line tangent to the semicircle, the -axis and Now, we introduce the value inverse to by and the radius of the circle remains to be right angled. This is rewrite (17)intheform exactly the case considered in Section 2.2,consequently,we 𝑎 𝐵 can apply formula (13). According to (13)wewrite exp (− )=tan . (19) K 2 𝑟 𝑟 𝐴𝐵 =𝑟 coth (𝑟𝜙) = ,𝐴𝐶1 = =𝑟tan 𝐵. Let the point 𝐴 run away from the point 𝐶 to infinity. The cos 𝐵 sinh (𝑟𝜙) following two cases can be considered. (22) (1) 𝜙 K tends to zero, and tends to infinity; then the angle Since 󳵻𝐴𝐵𝐶1 ∼󳵻𝐴𝐾1𝑃1,wehave 𝐵 will tend to right angle. 𝑃1𝐾1 𝑟 ThisistrueinEuclideanplane. = = sinh (𝑟𝜙) , (23) 𝐴𝐾1 𝐴𝐶1 (2) Suppose that 𝜙,andK and the angle 𝐵 go to some limited values, 𝑃1𝐾1 =𝐴𝐾1 sinh (𝑟𝜙) = (𝐴𝐵 −𝑟) sinh (𝑟𝜙) (24) lim K =𝜅, lim 𝐵=Π(𝑎) . 𝐴𝐶󳨀→∞ (20) =𝑟exp (−𝑟𝜙)2 ,𝑃 𝐾2 =𝑟exp (𝑟𝜙) . 4 Journal of Mathematics

On the basis of obtained formulae the following relationships to use formula (27). Consider arc ⌣𝐶1𝐶2 isdefinedinthe between circular and hyperbolic trigonometric functions are first quadrant of the semicircle with end-points at 𝐶1 and 𝐶2. established: The arc ⌣𝐶1𝐶2 can be presented as difference of two arcs, 1 both originated from the top of the semicircle: (𝐵) = , (𝐵) = (𝑟𝜙) , sin (𝑟𝜙) cot sinh cosh (25) ⌣𝐶1𝐶2 =⌣ 𝑁𝐶1−⌣𝑁𝐶2. (30) cos (𝐵) = tanh (𝑟𝜙) . Denote by 𝑎1, 𝑎2 the angles formed by radiuses 𝐵𝐶1 and 𝐵𝐶2 with 𝑥-axis, correspondingly, where 𝐵 is a center of the If we make the point 𝐶 tend to the top of semicircle 𝑁,the circle. Denote the hyperbolic angle corresponding to the arcs hyperbolic angle 𝜙 will tend to zero. When the point 𝐶 tends ⌣𝑁𝐶𝑘, 𝑘=1,2by 𝜉(𝑁𝐶𝑘), 𝑘=1,2.Thenthefunctions to end-point 𝐾1, the hyperbolic angle tends to infinity. Thus, the hyperbolic angle is measured from the point 𝑁,thetopof cosh 𝜉(𝑁𝐶1), cosh 𝜉(𝑁𝐶2), the semicircle. Consider two different positions of the tangent (31) line, corresponding to two positions 𝐶1, 𝐶2, with hyperbolic sinh 𝜉(𝑁𝐶1), sinh 𝜉(𝑁𝐶2) angles 𝜙2 and 𝜙1. According to key formula (6)wewrite are expressed via circular trigonometric functions according 𝑥2 −𝐷2 exp ((𝑥2 −𝑥1)𝜙2)= , to formulae (25): 𝑥1 −𝐷2 (26) 1 1 𝑥 −𝐷 𝜉(𝑁𝐶 )= , 𝜉(𝑁𝐶 )= , ((𝑥 −𝑥 )𝜙 )= 2 1 . cosh 1 cosh 2 exp 2 1 1 sin 𝑎1 sin 𝑎2 𝑥1 −𝐷1 (32) sinh 𝜉(𝑁𝐶1)=cot 𝑎1, sinh 𝜉(𝑁𝐶2)=cot 𝑎2. The hyperbolic angle 𝜙2 corresponds to the arc ⌣𝑁𝐶2,and 𝜙 ⌣𝑁𝐶 the hyperbolic angle 1 corresponds to the arc 1.Then Then, by taking into account additional formulae wesupposethatthedifferenceinthehyperbolicangles𝜙2 −𝜙1 will correspond to the arc ⌣𝐶2𝐶1.From(26)itfollowsthat cosh 𝜉(𝐶1𝐶2)= cosh 𝜉(𝑁𝐶1) cosh 𝜉(𝑁𝐶2)

𝑥2 −𝐷2 𝑥1 −𝐷1 (𝑥 −𝑥 )(𝜙 −𝜙 )= , − sinh 𝜉(𝑁𝐶1) sinh 𝜉(𝑁𝐶2), exp 2 1 2 1 𝑥 −𝐷 𝑥 −𝐷 1 2 2 1 (27) (33) sinh 𝜉(𝐶1𝐶2)= sinh 𝜉(𝑁𝐶1) cosh 𝜉(𝑁𝐶2) 𝐷𝑘 =𝐷(𝜙𝑘) , 𝑘 = 1, 2. − cosh 𝜉(𝑁𝐶1) sinh 𝜉(𝑁𝐶2), 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 we get hyperbolic trigonometric functions of the arcs with geodesic line in the Poincaremodel.Let´ 𝑧, 𝑤 ∈𝐻, and let arbitrary end-points on the semicircle expressed via periodic 𝑧 𝑤 𝑧 1 and 1 be end points of a geodesic line passing through trigonometric functions: and 𝑤. Then the distance between these points is defined by formula 1− 𝑎 𝑎 𝜉(𝐶 𝐶 )= cos 1 cos 2 , 𝜌(𝑧 ,𝑧 )= (𝑧 ,𝑧 ;𝑧 ,𝑧 ), cosh 1 2 1 3 log 1 2 3 4 (28) sin 𝑎1 sin 𝑎2 (34a) 𝑎 − 𝑎 where 𝜉(𝐶 𝐶 )=cos 1 cos 2 . sinh 1 2 𝑎 𝑎 𝑧 −𝑧 𝑧 −𝑧 sin 1 sin 2 (𝑧 ,𝑧 ;𝑧 ,𝑧 ):= 2 1 4 3 1 2 3 4 𝑧 −𝑧 𝑧 −𝑧 (29) 4 1 2 3 Compare with analogous formula from Poincaremodel([´ 3], formula (1.2.6)) given by is the cross-ratio. Comparing (29)with(27)wecometothe 𝑧 =𝑥 𝑧 =𝑥 𝑧 =𝐷 following correspondence: 2 2, 4 1, 1 2, |𝑧−𝑤|2 𝑧3 =𝐷1. Notice, however, in our construction 𝐷𝑘, 𝑘=1,2 cosh 𝜌 (𝑧,) 𝑤 =1+ . (34b) 2 Im (𝑧) Im (𝑤) 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. Two complex numbers

3. Relationships between Elements of Three 𝑧=V0 +𝑖V,𝑤=𝑤0 +𝑖𝑤 (35) Intersecting Semicircles are related to the radius of the semicircle and the angles 𝑎1,𝑎2 3.1. Hyperbolic Cosine-Sine Functions of Arcs of the Semicircle. by The key formula (6) admits to define hyperbolic trigonomet- ric functions of the arcs originated from the top 𝑁 of the V =𝑟sin 𝑎2, V0 =𝑟cos 𝑎2, semicircle C. In order to determine trigonometric functions (36) 𝑤=𝑟 𝑎 ,𝑤=𝑟 𝑎 . of the arcs with arbitrary end-points on the semicircle we have sin 1 0 sin 1 Journal of Mathematics 5

𝐴 Then, 𝐶

𝐵 cos 𝛼=cos 𝑏1 cos 𝑐1 + sin 𝑏1 sin 𝑐1, (40a)

cos 𝛽=cos 𝑎2 cos 𝑐2 − sin 𝑎2 sin 𝑐2, (40b)

𝑂𝑐 𝑂𝐵ℎ𝐴 ℎ𝐵 ℎ𝐶 𝑂𝑎 cos 𝛿=−cos 𝑏2 cos 𝑎1 + sin 𝑏2 sin 𝑎1, (40c) Figure 3: Triangle formed by arcs of intersecting semicircles. sin 𝛼=sin 𝑏1 cos 𝑐1 − cos 𝑏1 sin 𝑐1, (41a)

sin 𝛽=sin 𝑎2 cos 𝑐2 + cos 𝑎2 sin 𝑐2, (41b) Then, 𝛿= 𝑏 𝑎 + 𝑏 𝑎 . 2 sin sin 2 cos 1 cos 2 sin 1 (41c) 1−cos 𝑎1 cos 𝑎2 𝑟 − V0𝑤0 cosh 𝜉(𝐶1𝐶2)= = sin 𝑎1 sin 𝑎2 𝑤V Denote distances between centers by (V −𝑤 )2 + (V −𝑤)2 𝑂 =𝑂𝑂 ,𝑂=𝑂𝑂 ,𝑂=𝑂𝑂 . =1+ 0 0 = 𝜌 (𝑧,) 𝑤 . 𝑐𝑏 𝑐 𝑏 𝑏𝑎 𝑏 𝑎 𝑎𝑐 𝑎 𝑐 (42) 2𝑤V cosh (37) The theorem of sines employed for triangles 𝑂𝑐𝐴𝑂𝑏, 𝑂𝑏𝐶𝑂𝑎, 𝑂𝑐𝐵𝑂𝑎 gives six relations of type 3.2. Relationships between Elements of the Triangle Bounded by Arcs of the Intersecting Semicircles. In Figure 3 three sin 𝛼 sin 𝑐1 sin 𝑏1 intersecting semicircles with centers installed on horizontal = = , (43a) 𝑂𝑐𝑏 𝑟𝑏 𝑟𝑐 axis at the points 𝑂𝑎, 𝑂𝑏,and𝑂𝑐 are presented. Intersections 󳵻𝐴𝐵𝐶̃ ⌣ of the semicircles form triangle bounded by the arcs sin 𝛿 sin 𝑎1 sin 𝑏2 𝐵𝐶, ⌣ 𝐴𝐵, ⌣𝐴𝐶 = = , . For each arc we can put in correspondence 𝑂 𝑟 𝑟 (43b) the hyperbolic angle. Denote the hyperbolic angles by 𝑎, 𝑏,, 𝑐 𝑏𝑎 𝑏 𝑎 𝑎, 𝑏, 𝑐 ⌣ where consecutively will correspond to the arcs sin 𝛽 sin 𝑐2 sin 𝑎2 𝐵𝐶, ⌣ 𝐴𝐶, ⌣𝐴𝐵 = = . (43c) . 𝑂𝑎𝑐 𝑟𝑎 𝑟𝑐 Connect vertices of the triangle with centers of the circle 𝑎 𝑏 𝑐 𝑘=1,2 by corresponding radiuses. Denote by 𝑘, 𝑘, 𝑘, the From these relations the first set of main relationships follows. angles bounded by the radiuses and 𝑥-axis, where 𝑎1 >𝑎2 >0, 𝑏1 >𝑏2 >0, 𝑐1 >𝑐2 >0.Bymakinguseof(34a) define Relation 1. Consider hyperbolic cosine-sine functions corresponding to bounding segments: 𝑟𝑎 sin 𝑎1 =𝑟𝑏 sin 𝑏2,𝑟𝑐 sin 𝑐2 =𝑟𝑎 sin 𝑎2, (44) 𝑟𝑐 sin 𝑐1 =𝑟𝑏 sin 𝑏1. 1−cos 𝑎1 cos 𝑎2 cos 𝑎1 − cos 𝑎2 cosh 𝑎= , sinh 𝑎= , sin 𝑎1 sin 𝑎2 sin 𝑎1 sin 𝑎2 From the draught in Figure 3 it is seen that (38a) 𝑂𝑎𝑐 =𝑂𝑏𝑎 +𝑂𝑐𝑏, (45) 1−cos 𝑏1 cos 𝑏2 cos 𝑏1 − cos 𝑏2 cosh 𝑏= , sinh 𝑏= , sin 𝑏1 sin 𝑏2 sin 𝑏1 sin 𝑏2 where (38b) 𝑂𝑎𝑐 =𝑟𝑐 cos 𝑐2 +𝑟𝑎 cos 𝑎2,𝑂𝑏𝑎 =𝑟𝑎 cos 𝑎1 +𝑟𝑏 cos 𝑏2. 1−cos 𝑐1 cos 𝑐2 cos 𝑐1 − cos 𝑐2 cosh 𝑐= , sinh 𝑐= . (46) sin 𝑐1 sin 𝑐2 sin 𝑐1 sin 𝑐2 (38c) Hence,

Theusualnotionoftheangleisusedthatis,theangle 𝑂𝑐𝑏 =𝑟𝑐 cos 𝑐2 +𝑟𝑎 cos 𝑎2 −𝑟𝑎 cos 𝑎1 −𝑟𝑏 cos 𝑏2. (47) between two curves is defined as an angle between their tangent lines. Let the angles 𝛼, 𝛽, 𝛿 be angles at the verteices From vertices of 󳵻𝐴𝐵𝐶̃ erect lines perpendicular to horizon- 𝐴, 𝐵, 𝐶 , correspondingly. For these angles we can define thier tal line intersect with 𝑥-axis at points ℎ𝐴,ℎ𝐵,ℎ𝐶, correspond- proper cosine and sine functions. The angles of the triangle ingly. From Figure 3 we find that ̃ 󳵻𝐴𝐵𝐶 𝛼, 𝛽,𝛿 arecloselyrelatedtoangles𝑎1,𝑎2,𝑏1,𝑏2,𝑐1,𝑐2. From Figure 3 we find the following relationships between 𝑂𝑐𝑏 =𝑂𝑐ℎ𝐴 −ℎ𝐴𝑂𝑏 =𝑟𝑐 cos 𝑐1 −𝑟𝑏 cos 𝑏1. (48) them: By equating (47)with(48) we arrive to another main 𝛽=𝑎2 +𝑐2,𝛿=𝜋−𝑎1 −𝑏2,𝛼=𝑏1 −𝑐1. (39) relationship between radii and angles. 6 Journal of Mathematics

Relation 2. One has Denote the segments projections of sides of 󳵻𝐴𝐵𝐶̃ on 𝑥-axis by P(𝐴𝐶)𝐴 =ℎ ℎ𝐶, P(𝐴𝐵)𝐴 =ℎ ℎ𝐵, P(𝐵𝐶)𝐵 =ℎ ℎ𝐶.From 𝑟𝑐 cos 𝑐1 −𝑟𝑏 cos 𝑏1 =𝑟𝑐 cos 𝑐2 +𝑟𝑎 cos 𝑎2 −𝑟𝑎 cos 𝑎1 (49) Figure 3 it is seen that −𝑟 𝑏 . 𝑏 cos 2 P (𝐴𝐶) = P (𝐴𝐵) + P (𝐵𝐶) , (57) We will effect a simplification by using the following where designations: P (𝐵𝐶) =𝑤01 − V01, P (𝐴𝐶) =𝑤02 − V02, 𝑟𝑎 cos 𝑎1 =𝑤01,𝑟𝑎 sin 𝑎1 =𝑤1, (58) P (𝐴𝐵) =𝑤03 − V03. 𝑟𝑎 cos 𝑎2 = V01,𝑟𝑎 sin 𝑎2 = V1, 4. Hyperbolic Law of Cosines I for the Triangle 𝑟𝑏 cos 𝑏1 =𝑤02,𝑟𝑏 sin 𝑏1 =𝑤2, (50) Formed by Intersection of Three Semicircles 𝑟𝑏 cos 𝑏2 = V02,𝑟𝑏 sin 𝑏2 = V2, The main aim of this section is to prove the hyperbolic law 𝑟 𝑐 =𝑤 ,𝑟𝑐 =𝑤, 𝑐 cos 1 03 𝑐 sin 1 3 theorem of cosines I for the triangle 󳵻𝐴𝐵𝐶̃ formed by arcs of intersecting semicircles with centers installed on 𝑥-axis 𝑟𝑐 cos 𝑐2 = V03,𝑟𝑐 sin 𝑐2 = V3. (Figure 3). This law is given by the following set of equations: In these designations formulae (40a), (40b), (40c)and(41a), cosh 𝑐=cosh 𝑎 cosh 𝑏−sinh 𝑎 sinh 𝑏 cos 𝛿, (41b), (41c) are written as follows: cosh 𝑏=cosh 𝑎 cosh 𝑐−sinh 𝑐 sinh 𝑎 cos 𝛽, (59) 𝑟𝑏𝑟𝑐 sin 𝛼=𝑤2𝑤03 −𝑤02𝑤3, (51a) cosh 𝑎=cosh 𝑐 cosh 𝑏−sinh 𝑐 sinh 𝑏 cos 𝛼. 𝑟𝑏𝑟𝑐 cos 𝛼=𝑤02𝑤03 +𝑤2𝑤3, Theorem 3 (theorem of cosines I). The following equation 𝑟 𝑟 𝛽=V V + V V , 𝑎 𝑐 sin 1 03 01 3 for elements of the triangle 󳵻𝐴𝐵𝐶̃ formed by arcs of three (51b) intersecting semicircles holds true 𝑟𝑎𝑟𝑐 cos 𝛽=V01V03 − V1V3, cosh 𝑐=cosh 𝑎 cosh 𝑏−sinh 𝑎 sinh 𝑏 cos 𝛿, (60) 𝑟𝑎𝑟𝑏 sin 𝛿=V2𝑤01 + V02𝑤1, (51c) 𝑐, 𝑎, 𝑏, 𝑎, 𝑏, 𝛿 𝑟 𝑟 𝛿=V 𝑤 − V 𝑤 . where cosh cosh cosh sinh sinh cos are defined by 𝑎 𝑏 cos 2 1 02 01 formulae (51a), (51b), (51c)-(52). Formulae (38a), (38b), (38c) for hyperbolic sines and cosines Proof. Square both sides of Relation 2 to obtain are rewritten as follows: 2 2 2 2 (𝑤 − V ) =(𝑤 − V ) +(𝑤 − V ) 𝑟 −𝑤01V01 𝑤 − V 03 03 02 02 01 01 𝑎= 𝑎 , 𝑎=𝑟 01 01 , (61) cosh 𝑤 V sinh 𝑎 𝑤 V 1 1 1 1 −2(𝑤02 − V02)(𝑤01 − V01) 2 𝑟𝑏 −𝑤02V02 𝑤02 − V02 andevaluatethisequalitybytakingintoaccountformulae cosh 𝑏= , sinh 𝑏=𝑟𝑏 , (52) 𝑤2V2 𝑤2V2 (51a), (51b), (51c)-(52).Firstofall,evaluatetheleft-handside of this equation as follows: 2 𝑟 −𝑤03V03 𝑤 − V 𝑐= 𝑐 , 𝑐=𝑟 03 03 . (𝑤 − V )2 = V2 +𝑤2 −2V 𝑤 cosh 𝑤 V sinh 𝑐 𝑤 V 03 03 03 03 03 03 3 3 3 3 (62) 2 2 2 2 In the designations equations of Relation 1 are written as = V03 +𝑤03 −2𝑟𝑐 +2(𝑟𝑐 − V03𝑤03),

𝑥:=𝑤2 =𝑤3,𝑦:=V1 = V3,𝑧:=𝑤1 = V2. (53) where 2𝑟2 =𝑤2 +𝑤2 + V2 + V2. Equations (43a), (43b), (43c)–(46) are rewritten as follows: 𝑐 03 3 03 3 (63) Use (63)andwrite(62) as follows: 𝑂𝑐𝑏 =𝑤03 −𝑤02,𝑂𝑎𝑐 = V03 + V01, (54) 2 2 2 2 2 2 2 2 2 V03 +𝑤03 −2𝑟𝑐 = V03 +𝑤03 −(𝑤03 +𝑤3 + V03 + V3) 𝑂𝑏𝑎 =𝑤01 + V02. (64) 2 2 Correspondingly, Relation 2 takes the form =−(𝑤3 + V3).

𝑤03 −𝑤02 = V03 + V01 −𝑤01 − V02. (55) Equation (61) takes the following form: 2 2 2 2 2 This expresses the fact that 𝑂𝑐𝑎 is a sum of 𝑂𝑐𝑏 and 𝑂𝑏𝑎.Notice −(𝑤⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟3 + V3) +2(𝑟𝑐 − V03𝑤03)=(𝑤02 − V02) +(𝑤01 − V01) that (55)canberewrittenalsoinanotherequivalentform namely, −2(𝑤02 − V02)(𝑤01 − V01). 𝑤03 − V03 =𝑤02 − V02 − (𝑤01 − V01) . (56) (65) Journal of Mathematics 7

The underlined term passes to the right-hand side of the Replace the underlined term of (69)by(72), and pass equation. Then in the left-hand side remains the expression expression containing sines and cosines to the left-hand side oftheobtainedequation.Asaresult,wecometothefollowing 2(𝑟2 − V 𝑤 )=2V 𝑤 𝑐. 𝑐 03 03 3 3 cosh (66) equation: V 𝑤 Dividing both sides of the obtained equation by 3 3,weget 2 cosh 𝑐−2cos 𝛿 sinh 𝑏 sinh 𝑎

1 2 2 2 2 1 2 2 2 2 cosh 𝑐= (𝑤3 + V3 +(V02 −𝑤02) +(V01 −𝑤01) = {V 𝑤 (𝑤 + V +(𝑤 − V ) V3𝑤3 2 1 2 1 02 02 𝑤2V2𝑤1V1 (73) 2 −2 (V02 −𝑤02)(V01 −𝑤01)), +(𝑤01 − V01) )−2V02𝑤01 (67) ×(𝑤02 −V02)(𝑤01 −V01)} according to Relation 1 𝑤3 =𝑤2, V3 = V1.Theserelations make true the following equation By applying the elementary algebra one may show that (see, 1 1 1 [10]) = V2𝑤1. (68) 𝑤3V3 𝑤2V2 𝑤1V1 2 2 2 2 V2𝑤1 (𝑤2 + V1 +(𝑤02 − V02) +(𝑤01 − V01) ) On making use this formula in the right-hand side of (67)we come to the following equation: −2V02𝑤01 (𝑤02 − V02)(𝑤01 − V01) (74) 1 1 =2(𝑟2 − V 𝑤 )(𝑟2 − V 𝑤 ). 2 cosh 𝑐= 𝑎 01 01 𝑏 02 02 𝑤2V2 𝑤1V1

2 2 2 2 Thus, in the right-hand side73 of( )wehave ×{V2𝑤1 (𝑤3 + V3 +(V02 −𝑤02) +(V01 −𝑤01) )} 1 2 2 1 1 2 (𝑟 − V 𝑤 )(𝑟 − V 𝑤 ) , 𝑤 V V 𝑤 𝑎 01 01 𝑏 02 02 (75) −2 V2𝑤1 (V02 −𝑤02)(V01 −𝑤01). 1 1 2 2 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟𝑤2V2 𝑤1V1 which according to formulae (52)isnothingelsethan (69) 2 cosh 𝑎 cosh 𝑏. Evaluate now the underlined term, which we firstly write as (76) follows: By using it in (73) we arrive to the following equation: 1 1 2(V02 −𝑤02)(V01 −𝑤01) V2𝑤1 2 𝑐−2 𝑎 𝑏=2 𝛿 𝑎 𝑏. 𝑤2V2 𝑤1V1 cosh cosh cosh cos sinh sinh (77) (70) V 𝑤 𝑟 (V −𝑤 ) 𝑟 (V −𝑤 ) =2 2 1 𝑏 02 02 𝑎 01 01 . 𝑟 𝑟 𝑤 V 𝑤 V 𝑎 𝑏 2 2 1 1 5. Hyperbolic Laws of Sines and Cosines II From the second equation of (51c)wehave The main task of this section is to prove hyperbolic law V2𝑤1 V02𝑤01 (theorem) of sines,whichisgivenbytheformulae = cos 𝛿+ . (71) 𝑟𝑎𝑟𝑏 𝑟𝑎𝑟𝑏 sinh 𝑎 sinh 𝑏 sinh 𝑐 = = , (78) By making use of (71)in(70), evaluate (70) as follows: sin 𝛼 sin 𝛽 sin 𝛿 V 𝑤 𝑟 (V −𝑤 ) 𝑟 (V −𝑤 ) 2 2 1 𝑏 02 02 𝑎 01 01 and the hyperbolic law (theorem) of cosines II given by the 𝑟𝑎𝑟𝑏 𝑤2V2 𝑤1V1 formulae 𝑟 (V −𝑤 ) 𝑟 (V −𝑤 ) V02𝑤01 𝑏 02 02 𝑎 01 01 cos 𝛿=−cos 𝛼 cos 𝛽−sin 𝛼 sin 𝛽 cosh 𝑐, (79) =2(cos 𝛿+ ) 𝑟𝑎𝑟𝑏 𝑤2V2 𝑤1V1 cos 𝛽=−cos 𝛼 cos 𝛿−sin 𝛼 sin 𝛿 cosh 𝑏, (80) 𝑟𝑏 (V02 −𝑤02) 𝑟𝑎 (V01 −𝑤01) =2cos 𝛿 cos 𝛼=−cos 𝛽 cos 𝛿−sin 𝛿 sin 𝛽 cosh 𝑎. (81) 𝑤2V2 𝑤1V1 (V −𝑤 ) (V −𝑤 ) −2V 𝑤 02 02 01 01 5.1. Hyperbolic Theorem of Sines and Its Geometrical Interpre- 02 01 𝑤 V 𝑤 V 2 2 1 1 tation on Euclidean Plane (V −𝑤 ) (V −𝑤 ) =2 𝛿 𝑎 𝑏−2V 𝑤 02 02 01 01 . Theorem 4. Theratiosofprojectionsofthesidesoftriangle cos sinh sinh 02 01 ̃ 𝑤2V2 𝑤1V1 󳵻𝐴𝐵𝐶 on 𝑥-axis to corresponding distances between centers of (72) the semicircles are equal to each other. 8 Journal of Mathematics

̃ Proof. Projections of the sides of 󳵻𝐴𝐵𝐶 are given by formulae Having these formulae we may present the projection P𝑐𝑎 as follows P𝑏𝑐 =𝑟𝑎 cos 𝑎1 −𝑟𝑎 cos 𝑎2, P𝑐𝑎 =𝑟𝑏 cos 𝑏1 −𝑟𝑏 cos 𝑏2, P𝑎𝑐 =𝑟𝑏 cos 𝑏1 −𝑟𝑏 cos 𝑏2 P𝑎𝑏 =𝑟𝑐 cos 𝑐1 −𝑟𝑐 cos 𝑐2. (82) −𝑂2 +𝑟2 −𝑟2 𝑂2 −𝑟2 +𝑟2 = 𝑐𝑏 𝑐 𝑏 − 𝑏𝑎 𝑎 𝑏 2𝑂 2𝑂 Distances between centers of the circles have been defined as 𝑐𝑏 𝑏𝑎 (90) (see, (46)and(47)) 2 2 2 2 2 2 −𝑂𝑐𝑏𝑂𝑏𝑎 +𝑟𝑐 𝑂𝑏𝑎 −𝑟𝑏 𝑂𝑏𝑎 −𝑂𝑏𝑎𝑂𝑐𝑏 +𝑟𝑎𝑂𝑐𝑏 −𝑟𝑏 𝑂𝑐𝑏 𝑂 =𝑟 𝑐 +𝑟 𝑎 ,𝑂=𝑟 𝑎 +𝑟 𝑏 , = 𝑐𝑎 𝑐 cos 2 𝑎 cos 2 𝑏𝑎 𝑎 cos 1 𝑏 cos 2 2𝑂𝑐𝑏𝑂𝑏𝑎 𝑂 =𝑟 𝑐 −𝑟 𝑏 , 𝑐𝑏 𝑐 cos 1 𝑏 cos 1 The first ratio is presented as follows: (83) P −𝑂 𝑂 (𝑂 +𝑂 )+𝑟2𝑂 −𝑟2 (𝑂 −𝑂 )+𝑟2𝑂 𝑐𝑎 = 𝑐𝑏 𝑏𝑎 𝑐𝑏 𝑏𝑎 𝑐 𝑏𝑎 𝑏 𝑏𝑎 𝑐𝑏 𝑎 𝑐𝑏 P𝑐𝑎 = P𝑏𝑐 + P𝑎𝑏,𝑂𝑐𝑎 =𝑂𝑏𝑐 +𝑂𝑎𝑏. (84) 𝑂𝑐𝑎 2𝑂𝑐𝑏𝑂𝑏𝑎𝑂𝑐𝑎 Relation 1 given by the set of equations −𝑂 𝑂 (𝑂 )+𝑟2𝑂 −𝑟2 (𝑂 )+𝑟2𝑂 = 𝑐𝑏 𝑏𝑎 𝑐𝑎 𝑐 𝑏𝑎 𝑏 𝑐𝑎 𝑎 𝑐𝑏 . 𝑟𝑎 sin 𝑎1 =𝑟𝑏 sin 𝑏2,𝑟𝑐 sin 𝑐2 =𝑟𝑎 sin 𝑎2, 2𝑂 𝑂 𝑂 (85) 𝑐𝑏 𝑏𝑎 𝑐𝑎 (91) 𝑟𝑐 sin 𝑐1 =𝑟𝑏 sin 𝑏1 raises to the second power, Now, in the same way let us calculate the next ratio, namely, P /𝑂 2 2 2 2 2 2 𝑏𝑐 𝑏𝑐. Formula for the projection is evaluated as follows: 𝑟𝑎 −𝑟𝑎cos 𝑎1 =𝑟𝑏 −𝑟𝑏 cos 𝑏2, P =𝑟 𝑎 −𝑟 𝑎 2 2 2 2 2 2 𝑏𝑐 𝑎 cos 1 𝑎 cos 2 𝑟𝑐 −𝑟𝑐 cos 𝑐2 =𝑟𝑎 −𝑟𝑎cos 𝑎2, (86) 𝑂2 −𝑟2 +𝑟2 𝑂2 −𝑟2 +𝑟2 2 2 2 2 2 2 = 𝑏𝑎 𝑏 𝑎 − 𝑐𝑎 𝑐 𝑎 𝑟 −𝑟 𝑐1 =𝑟 −𝑟 𝑏1. 𝑐 𝑐 cos 𝑏 𝑏 cos 2𝑂𝑏𝑎 2𝑂𝑎𝑐 2 Then, square distances 𝑂𝑖𝑘,𝑖,𝑘=𝑎,𝑏,𝑐and use (86). We get 2 2 2 2 ((𝑂 𝑏𝑎𝑂𝑎𝑐(𝑂𝑏𝑎 −𝑂𝑐𝑎)−𝑟𝑏 𝑂𝑎𝑐 +𝑟𝑎𝑂𝑎𝑐)+𝑟𝑐 𝑂𝑏𝑎 −𝑟𝑎𝑂𝑏𝑎) 2 2 2 2 2 = . 𝑂𝑐𝑎 =𝑟𝑐 cos 𝑐2 +𝑟𝑎cos 𝑎2 +2𝑟𝑐𝑟𝑎 cos 𝑐2 cos 𝑎2, 2𝑂𝑎𝑐𝑂𝑏𝑎 2 2 2 2 2 (92) 𝑂𝑏𝑎 =𝑟𝑎cos 𝑎1 +𝑟𝑏 cos 𝑏2 +2𝑟𝑎𝑟𝑏 cos 𝑎1 cos 𝑏2, (87) −𝑂 =𝑂 −𝑂 2 2 2 2 2 By taking into account equation 𝑐𝑏 𝑏𝑎 𝑎𝑐,weget 𝑂𝑐𝑏 =𝑟𝑐 cos 𝑐1 +𝑟𝑏 cos 𝑏1 −2𝑟𝑐𝑟𝑏 cos 𝑐1 cos 𝑏1. 2 2 2 (𝑂𝑏𝑎𝑂𝑎𝑐𝑂𝑐𝑏 −𝑟𝑏 𝑂𝑎𝑐 +𝑟𝑎𝑂𝑐𝑏 +𝑟𝑐 𝑂𝑏𝑎) Combine (86)with(87); in this way we come to the following P = . (93) 𝑏𝑐 2𝑂 𝑂 system of equations: 𝑎𝑐 𝑏𝑎 2 2 2 (𝑎) 𝑂𝑐𝑎 =𝑟𝑐 −𝑟𝑎 +2𝑟𝑎 cos 𝑎2𝑂𝑎𝑐, Next, let us calculate the ratio (88a) 2 2 2 2 2 2 (𝑏) 𝑂 =𝑟 −𝑟 +2𝑟 𝑐 𝑂 , P (𝑂𝑏𝑎𝑂𝑎𝑐 (𝑂𝑐𝑏)−𝑟𝑏 𝑂𝑎𝑐 +𝑟𝑎 (𝑂𝑐𝑏)+𝑟𝑐 𝑂𝑏𝑎) 𝑐𝑎 𝑎 𝑐 𝑐 cos 2 𝑎𝑐 𝑏𝑐 = . (94) 𝑂 2𝑂 𝑂 𝑂 2 2 2 𝑏𝑐 𝑎𝑐 𝑏𝑎 𝑏𝑐 (𝑎) 𝑂𝑏𝑎 =𝑟𝑎 −𝑟𝑏 +2𝑟𝑏 cos 𝑏2𝑂𝑏𝑎, (88b) 2 2 2 This expression coincides with91 ( ); hence, (𝑏) 𝑂𝑏𝑎 =𝑟𝑏 −𝑟𝑎 +2𝑟𝑎 cos 𝑎1𝑂𝑏𝑎, P𝑏𝑐 P𝑎𝑐 2 2 2 = . (95) (𝑎) 𝑂𝑐𝑏 =𝑟𝑏 −𝑟𝑐 +2𝑟𝑐 cos 𝑐1𝑂𝑐𝑏, 𝑂𝑏𝑐 𝑂𝑎𝑐 (88c) 2 2 2 (𝑏) 𝑂𝑐𝑏 =𝑟𝑐 −𝑟𝑏 −2𝑟𝑏 cos 𝑏1𝑂𝑐𝑏. By taking into account (84), we arrive to desired relations

From these equations the cosines of the angles 𝑎1,𝑎2,𝑏1,𝑏2, P𝑏𝑐 P𝑎𝑐 P𝑎𝑏 𝑐 ,𝑐 = = . (96) 1 2 are expressed: 𝑂𝑏𝑐 𝑂𝑎𝑐 𝑂𝑎𝑏 𝑂2 −𝑟2 +𝑟2 𝑂2 −𝑟2 +𝑟2 𝑏𝑎 𝑏 𝑎 = 𝑎 , 𝑐𝑎 𝑐 𝑎 = 𝑎 , 2𝑟 𝑂 cos 1 2𝑟 𝑂 cos 2 Inthesequelcomebacktodesignationsintroducedin 𝑎 𝑏𝑎 𝑎 𝑎𝑐 Section 3. In these designations (96) is written as follows: 2 2 2 2 2 2 𝑂𝑐𝑏 −𝑟𝑏 +𝑟𝑐 𝑂𝑐𝑎 −𝑟𝑎 +𝑟𝑐 𝑤 − V 𝑤 − V 𝑤 − V = 𝑐 , = 𝑐 , (89) 01 01 02 02 03 03 2𝑟 𝑂 cos 1 2𝑟 𝑂 cos 2 = = . (97) 𝑐 𝑐𝑏 𝑐 𝑎𝑐 𝑤03 −𝑤02 V03 + V01 𝑤01 + V02 2 2 2 2 2 2 −𝑂𝑐𝑏 +𝑟𝑐 −𝑟𝑏 𝑂𝑏𝑎 −𝑟𝑎 +𝑟𝑏 Theorem 5. 󳵻𝐴𝐵𝐶̃ = cos 𝑏1, = cos 𝑏2. Thesidesandtheanglesoftriangle satisfy 2𝑟𝑏𝑂𝑐𝑏 2𝑟𝑏𝑂𝑏𝑎 (78). Journal of Mathematics 9

Proof. By using formulas (38a), (38b), and (38c)forsinh𝑎, We obtain 𝑏 𝑐 𝛼 sinh ,sinh and formulas (41a), (41b), and (41c)forsin , 1 2 2 2 1 2 2 2 sin 𝛽,sin𝛿,wewrite: cos 𝛼 cos 𝛽= (𝑟𝑐 +𝑟𝑏 −𝑂𝑐𝑏) (𝑟𝑐 +𝑟𝑎 −𝑂𝑎𝑐) 2𝑟𝑏𝑟𝑐 2𝑟𝑎𝑟𝑐 𝑟 (𝑤 − V ) 𝑥(𝑤 −𝑤 ) 1 sinh 𝑎 𝑎 01 01 03 02 = = : 4𝑟 𝑟 𝑟2 sin 𝛼 𝑦𝑧 𝑟𝑏𝑟𝑐 𝑏 𝑎 𝑐 𝑤 − V 𝑥𝑦𝑧 ×(𝑂2 𝑂2 −(𝑂2 +𝑂2 )𝑟2 −𝑂2 𝑟2 −𝑂2 𝑟2). = 01 01 , 𝑐𝑏 𝑎𝑐 𝑐𝑏 𝑎𝑐 𝑐 𝑐𝑏 𝑎 𝑎𝑐 𝑏 𝑤03 −𝑤02 𝑟𝑎𝑟𝑏𝑟𝑐 (102) 𝑏 𝑤 − V 𝑦(V + V ) 𝑤 − V 𝑥𝑦𝑧 sinh = 02 02 : 03 01 = 02 02 , By using (100)and(102) calculate the difference sin 𝛽 𝑟𝑏𝑥𝑧 𝑟𝑎𝑟𝑐 V03 + V01 𝑟𝑎𝑟𝑏𝑟𝑐 sin 𝛼 sin 𝛽 cosh 𝑐−cos 𝛼 cos 𝛽 𝑤 − V 𝑧(𝑤 + V ) 𝑤 − V 𝑥𝑦𝑧 sinh 𝑐 03 03 01 02 03 03 1 = : = . = (4𝑂 𝑂 𝑟2 −𝑂2 𝑂2 −(𝑂2 +𝑂2 )𝑟2 sin 𝛿 𝑟𝑐𝑦𝑥 𝑟𝑏𝑟𝑎 𝑤01 + V02 𝑟𝑎𝑟𝑏𝑟𝑐 2 𝑎𝑐 𝑐𝑏 𝑐 𝑐𝑏 𝑎𝑐 𝑐𝑏 𝑎𝑐 𝑐 4𝑟𝑎𝑟𝑏𝑟𝑐 (98) 1 +𝑂2 𝑟2 +𝑂2 𝑟2)− 𝑐𝑏 𝑎 𝑎𝑐 𝑏 2 It is seen that these equations contain a common factor which 4𝑟𝑏𝑟𝑎𝑟𝑐 is symmetric with respect to 𝑎, 𝑏, 𝑐 and 𝑥, 𝑦,. 𝑧 Multiply all ×(+𝑂2 𝑂2 −(𝑂2 +𝑂2 )𝑟2 −𝑂2 𝑟2 −𝑂2 𝑟2) equations of (97) by this factor. We arrive to (78). 𝑐𝑏 𝑎𝑐 𝑐𝑏 𝑎𝑐 𝑐 𝑐𝑏 𝑎 𝑎𝑐 𝑏 1 Theorem 6. Thesidesandtheanglesoftriangle󳵻𝐴𝐵𝐶̃ satisfy = (𝑟2 +𝑟2 −𝑂2 )= 𝛿. 2𝑟 𝑟 𝑎 𝑏 𝑎𝑏 cos the following equation: 𝑏 𝑎 (103) 𝛿= 𝛼 𝛽 𝑐− 𝛼 𝛽. cos sin sin cosh cos cos (99) Thus, we got (99).

Proof. Evaluate the first term of the right-hand side of (99) The other two equations, (80)and(81), are proved by making use of (43a), (43b), and (43c): analogously.

1 𝛼 𝛽 𝑐= 𝑂 𝑂 (1 − 𝑐 𝑐 ) 6. Concluding Remarks sin sin cosh 𝑟 𝑟 𝑐𝑎 𝑐𝑏 cos 1 cos 2 𝑎 𝑏 Wehave proved three hyperbolic cosine-sine theorems for the 1 triangular formed by arcs of three intersecting semicircles by = 𝑂𝑐𝑎𝑂𝑐𝑏 𝑟𝑎𝑟𝑏 using only elements of the Euclidean geometry. This geomet- rical figure is one of the basic figures of the Poincaremodelof´ 𝑂2 −𝑟2 +𝑟2 𝑂2 −𝑟2 +𝑟2 ×(1− 𝑐𝑏 𝑏 𝑐 𝑐𝑎 𝑎 𝑐 ) hyperbolic geometry. Conventionally, in the textbooks on the 2𝑟𝑐𝑂𝑐𝑏 2𝑟𝑐𝑂𝑎𝑐 Poincare´ model of hyperbolic geometry these theorems are proved by employing invariance properties of the cross-ratio 1 1 with respect to Mobius¨ group transformations. In that case = 𝑂𝑎𝑐𝑂𝑐𝑏 − 𝑂𝑎𝑐𝑂𝑐𝑏 𝑟𝑎𝑟𝑏 𝑟𝑎𝑟𝑏 one remains with the impression that the hyperbolic cosine- sine theorems are consequences of a special structure of the 𝐻 𝑂2 −𝑟2 +𝑟2 𝑂2 −𝑟2 +𝑟2 ×( 𝑐𝑏 𝑏 𝑐 𝑎𝑐 𝑎 𝑐 ) half-plane.Wehaveshownthatthehyperbolictrigonometry 2𝑟𝑐𝑂𝑐𝑏 2𝑟𝑐𝑂𝑎𝑐 as well as periodic trigonometry arises on Euclidean plane in anaturalway. 1 = (4𝑂 𝑂 𝑟2 −𝑂2 𝑂2 The method of parametrization of the mass-shell equa- 2 𝑎𝑐 𝑐𝑏 𝑐 𝑐𝑏 𝑐𝑎 4𝑟𝑎𝑟𝑏𝑟𝑐 tion via the hyperbolic angle, the rapidity,iswidelyusedinthe relativistic physics [4]. The quadratic equation (3)isclosely −(𝑂2 +𝑂2 )𝑟2 +𝑂2 𝑟2 +𝑂2 𝑟2). 𝑐𝑏 𝑐𝑎 𝑐 𝑐𝑏 𝑎 𝑎𝑐 𝑏 related to the mass-shell equation (100) 2 2 2 2 𝑝0 −𝑝 =𝑚𝑐 , (104) 𝛼 𝛽 Now calculate the product cos cos by using the following where 𝑚𝑐 plays the role of radius of the semicircle 2𝑚𝑐2 =𝑥 − formulae: 𝑥1. Within the framework of the present method we come to the following parametrization of energy momentum of the 1 𝛼= (𝑟2 +𝑟2 −𝑂2 ), relativistic particle [11]: cos 2𝑟 𝑟 𝑐 𝑏 𝑐𝑏 𝑏 𝑐 𝑚𝑐 (101) 𝑝0 =𝑚𝑐coth (𝑚𝑐𝜙) ,𝑝= , 0≤𝜙<∞. 1 2 2 2 sinh (𝑚𝑐𝜙) cos 𝛽=− (𝑟𝑐 +𝑟𝑎 −𝑂𝑎𝑐). 2𝑟𝑎𝑟𝑐 (105) 10 Journal of Mathematics

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).

References

[1] S. Stahl, The Poincare´ Half-Plane, Jones and Bartlett, Boston, Mass, USA, 1993. [2] J. W. Anderson, Hyperbolic Geometry, Springer, London, UK, 1999. [3] S. Katok, Fuchsian Groups, Chicago Lectures in Mathematics, University of Chicago Press, Chicago, Ill, USA, 1992. [4] A. A. Ungar, “Hyperbolic trigonometry and its application in the Poincareballmodelofhyperbolicgeometry,”´ Computers & Mathematics with Applications,vol.41,no.1-2,pp.135–147,2001. [5]R.M.Yamaleev,“Multicomplexalgebrasonpolynomialsand generalizedHamiltondynamics,”Journal of Mathematical Anal- ysis and Applications,vol.322,no.2,pp.815–824,2006. [6] R.M.Yamaleev,“Geometricalinterpretationofgeneralcomplex algebra and its application in relativistic mechanics,” Advances in Applied Clifford Algebras,vol.17,no.2,pp.281–305,2007. [7]V.F.Kagan,Foundations of Geometry. Part I., Gostekhizdat, Moscow, Russia, 1949. [8]N.J.Lobatschewsky,“Etudesgeom´ etriques´ sur la theorie´ de paralleles,` suivis d’un extrait de la correspondance de Gauss et de Schumacher,”in Memoires´ de la Societ´ e´ des sciences physiques et naturelles de Bordeaux,G.J.Hoauel,Ed.,vol.4,Soci¨ et´ edes´ sciences physiques et naturelles de Bordeaux, Bordeaux, France, 1866. [9] G. Dattoli and M. Del Franco, “Hyperbolic and circular trigonometryand application to special relativity,” In press, http://arxiv.org/abs/1002.4728 . [10] R. M. Yamaleev, “Hyperbolic cosines and sinestheorem for the triangle formed by intersections of three semicircles on Euclidean plane,” In press, http://arxiv.org/abs/1104.5135 . [11] R. M. Yamaleev, “New representation for energy-momentum and its applications to relativistic dynamics,” Physics of Atomic Nuclei,vol.74,no.12,pp.1775–1782,2011. [12] N. A. Chernikov, “Lobachevsky geometry and relativistic mechanics,” Physics of Atomic Nuclei,vol.4,no.3,pp.774–809, 1973. Advances in Advances in Journal of Journal of Operations Research Decision Sciences Applied Mathematics Algebra Probability and Statistics Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

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