SPHERICAL TRIGONOMETRY

Transcription

1 SPHERICAL TRIGONOMETRY THERE IS NO ROYAL ROAD TO GEOMETRY EUCLID BY KELLY LYNCH Saint May s College of Califonia Moaga 2016

2 Abstact The pupose of this pape is to deive vaious tigonometic fomulas fo spheical tiangles. The subject of spheical tigonomety has many navigational and astonomical applications. Histoy Geomety has been developing and evolving fo many centuies. Its uses ae vast and continue to affect ou evey day lives. The study of the sphee in paticula has its own unique stoy, and has two majo tuning points. This study fist began with the push of astonomy and was developed in depth by the Geeks. Thee is speculation that mathematical discoveies about the sphee wee made as ealy as the second centuy, but thee is no poof fo this.[1] The majo tansition fo the undestanding of spheical geomety was the wok of Menelaus of Alexanda. In his wok Spheica the autho delves deeply into the popeties of the sphee and the calculations elated to lengths and measues of a sphee s acs. Fo a long time the equations he discoveed, such as the measue of cicumfeence of a sphee and the measue of ac lengths, wee accepted and no futhe study was needed. The next majo motivation fo leaning spheical tigonomety was eligious mattes; the eligion of Islam equies that the diection of Mecca is always known fo daily paye. Menelaus findings wee futhe developed duing the Islamic Enlightenment peiod. Thee is some debate as to the discovey of the Law of Sines fo spheical tiangles. Possible souces of this discovey stem fom the debate ove the two Muslim scientists, Abū Nas o Abū ĺ-wafā. Who eve was esponsible fo the pogess in the Law of Sines allowed fo a moe concise poof to be developed late; as well as leading to othe theoems and identities on spheical tigonomety. Anothe majo name in geomety is Euclid. He made a big impact late in the thid centuy;

3 though the system of spheical tigonomety does not incopoate paallel lines, Euclidean geomety gave some insight to spheical behavio. In his wok Elements Euclid published equations which help lead us to the Pythagoean Theoem and the Law of Cosines. Though mathematicians bought insight to this aea of study, many influences on spheical tigonomety also came fom the field of science. Futhe discovey about the behavio of acs and angles became pominent in the late Renaissance peiod. John Napie, a Scottish scientist who lived aound the 17th centuy, was the fist to wok with ight spheical tiangles and the basic identities of these shapes. Using Napie s Rules, the law of cosines fo sphees was discoveed.[1] Definition On a sphee, a geat cicle is the intesection of the sphee with a plane passing though the cente, o oigin, of the sphee.[1] Example Geat Cicle

4 The solid line illustates a geat cicle which is the intesection of the plane X and the sphee A. The plane in the is the hoizontal plane; howeve, the plane can have any oientation that bisects the sphee. The cicle ceated by this intesection will have adius equal to the length of the adius of the sphee. It also follows that the length of the cicumfeence of the geat cicle will equal 2π whee is the length of the adius of the sphee. This is elevant because it enables us to calculate the length of a cicula segment by consideing the elation between the inne angle and the adius of the sphee. Theefoe, fo any sphee and any angle, the length of one ac segment will just equal θ whee θ is the measue of the angle in adians. This is tue because each cicula segment is just faction of the entie cicumfeence. We can show this elation of the angle and ac length in the example below. Example Geneal ac measue (whee a is the length of the ac This is a geneal example, but we can apply specific values as in the example on the following page.

5 Example Measue of Ac In this example we cut ou a segment of geat cicle by elating an inne angle. Fo this example let R = 4.69 and θ = adians, thus (4.69( = 4.39, which is the measue of the length of the ac. This example was ceated in a dynamic mathematics softwae pogam which gave these elated measuements. lune. The next definition equied in undestanding a spheical tiangle is that of a Definition Lune A lune is a pat of the sphee which is captued between two geat cicles.[1] This definition is elevant because it stated the ability to captue shapes on a sphee. This definition itself is not extemely significant, but it is though this shape which we can fom othe shapes on the sphee. The shape of a lune can also be seen in

6 the next figue. Example Lune Spheical tiangles can be defined in tems of lunes. Definition Spheical Tiangle A spheical tiangle is the intesection of thee distinct lunes.[1] In the figue above we can conside that thee ae two lunes which ae the on opposite sides of the sphee, it is natual that anothe lune bisecting these two will be needed. A simple way to think about a spheical tiangle is the shape captued though the intesection of thee geat cicles. It is also inteesting to conside that any two

7 unique points, which ae not diametic on the sphees suface, lie on a geat cicle. This makes sense because a tiangle would have thee vetices and theefoe thee diffeent geat cicles which go though two diffeent points. Since the spheical tiangle s edges ae cuved, it is clea that the equations, sides, aea, and geneal popeties will be diffeent fom that of a plana tiangle. We also know that in some way we should be able to elate the functions that we do discove to the adius of the sphee, because it seems natual that the fomulas would change in elation to the sphee s adius.we know that the behavio of a spheical tiangle will be vey diffeent fom the plana tiangles, thee ae even some definitions which can illustate this fo us. Example Spheical Tiangle Definition Spheical Excess is the amount by which the sum of the angles (in the spheical plane only exceed 180. This definition tells us about the behavio of the sphee and its edges. We know that the length of the edges on a spheical tiangle will be geate the edges on a coesponding plana tiangle, since they ae cuved. This definition allows fo a spheical tiangle to have multiple ight angles.

8 Example Spheical tiangle with thee ight angles Definition [3] Spheical Coodinates is a coodinate system in thee dimentions. The coodinate values stated below equie to be the length of the adius to the point P on the sphee. The value ϕ the angle between the z-axis, and the vecto fom the oigin to point P, and θ the angle between the x-axis, and the same vecto as in the figue Then we can say the x, y, z coodinates ae defined: os(θsin(ϕ, sin(θsin(ϕ, cos(ϕ Example Spheical Coodinates in the xyz plane

9 Theoem The Spheical Pythagoean Theoem[3] Fo a ight tiangle, ABC on a sphee of adius, with ight angle at vetex C and sides length a, b, c is defined: adius is 1. ( a ( b cos = cos cos This equation can equivalently be witten as cos(c = cos(acos(b when the Poof. Conside a spheical tiangle on the suface of a sphee with adius, and sides a, b, c. Let vetex C in the spheical tiangle be a ight angle, as in the diagam below.[3] Without loss of geneality, suppose C is defined by spheical coodinates C = (0, 0,. Now conside the othe vetices on the spheical tiangle. Call one of these A and let it be defined using spheical coodinates defined above A = ( sin ( b, 0, cos ( b whee b is equal to the measue of the ac length along the geat cicle between A and C. Note b is necessaily also equal the inteio angle between the vecto which goes fom the O to C and the vecto fom O to A, since the adius is equal to. Let the othe vetex called B be defined B = (0, sin ( a, cos ( a whee a is the ac length

10 between B to C, by the same easoning. Thus A = sin ( b ˆx + cos ( b ẑ and B = sin ( a ŷ + cos ( a ẑ. Now conside A B ( A B b = ( sin ˆx + cos ( b ( a = 2 cos cos ( b ( a ( a ẑ sin ŷ + cos ẑ Theefoesince the dot poduct of two vectos esults in a scala then A B = 2 cos ( b cos ( a. Now conside the magnitude of A B. Conside A B = A B cos A = since the adius is B = since the adius is A B = 2 cos Recall that the cosine of the angle between two vectos is just equal to the dot poduct of each vecto, ove the magnitude of each vecto. We know that the sign will not matte since the magnitude takes the absolute value so the signs will be equal, now using this infomation in elation to the equations above we can conclude that ( a ( b cos = cos cos

11 Theoem Law of Cosines The Law of Cosines states that if a,b,c ae the sides and A,B,C ae the angles of a spheical tiangle, then[1] ( a cos = cos cos ( b ( a + sin sin ( b cos(c Poof. Let thee be a spheical tiangle with sides denoted a, b, c. Let thei opposing angles be labeled A, B, C, whee A 90, B 90, and C 90. Let us call the cente of the sphee to be S. Theefoe we know fom any single vetex, A, B, C, to the point S is exactly equal to the adius,. If we abitay let A be the lagest angle(since we can abitaily name any angle then we can that thee is a point somewhee between the vetices B and C which ceates a ight angle, since A is the lagest angle and we must have spheical excess, it follows that ceating a ight angle is possible. Call this vetex N. This patition will ceate two ight spheical tiangles. Now we know that N some distance n fom B and is the distance (a n fom vetex C to N. Thee will also be some distance we shall call y fom the vetex N to point A. With this infomation we can use the spheical Pythagoean equation. So now applying the equation we ae left with two equations cos(b = cos(ncos(y cos(c = cos(a ncos(y Equation I Equation II

12 Now we can solve fo cos(y in equation I so theefoe we have: cos(y = cos(c cos(a n Now we need to solve using the same pocess fo equation II so we have: cos(y = cos(b cos(n cos(c cos(a n = cos(b cos(n cos(ccos(n = cos(bcos(a n combining equations coss multiply cos(ccos(n = cos(b(cos(acos(n + sin(asin(n cos(ccos(n = cos(bcos(acos(n + cos(bsin(asin(n cos(c = cos(bcos(a + sin(asin(b sin(n cos(n divide by cos(n cos(c = cos(acos(b + sin(acos(btan(n Napie 1 [1] cos(c = cos(acos(b + sin(asin(bcos(c which gives the law of cosines as desied. Also we can put this in tems of the adius of the sphee since we know that θ = x R whee x is the ac length and R is the adius of the sphee so if we substitute fo the sides and angles with espect to the adius we have 1 Napie discoveed the substitution fo this poof that cos(btan(n = sin(bcos(c

13 ( a cos = cos cos ( b ( a + sin sin ( b cos(c By equivalence we know that ( a cos = cos cos cos ( b + sin sin ( b cos(a ( b ( a ( a ( b = cos cos + sin sin cos(b This substitution can be made in any equation since the elation of inne angles to the ac length and will always be consistent fo that paticula sphee. Example Flight Routes Spheical tigonomety is vey applicable to ou evey day life. One of these applications is discoveing the length and diections needed fo navigation. The fomulas easily apply to the eath because the latitude and longitude lines ae actually examples of geat cicles. The law of cosines can be used on the example below, in which we want to find the distance between two cities. Given: Eath s adius: 6,371 km Distance fom San Fancisco, Califonia to Dublin, Ieland: 8198 km Distance fom San Fancisco, Califonia to Seattle, Washington: 1094 km

14 Angle between these two acs: (adians Now let us say we wish to find the distance fom Dublin to Seattle. To find this we use the equation below and then substitute the numeical values. ( ( a c = accos cos cos ( b ( a + sin sin ( b cos(c ( c = 6371 accos cos ( 1094 cos 6371 ( sin 6371 ( 1094 sin 6371 ( 8850 cos( c = 7232 km This value is an estimation because it assumes the eath is a pefect sphee with no dips o iegulaities which is not tue. This estimation is useful, howeve; since it was so quick and easy to find. Now we at least have some idea of the distance between two places that would othewise be impossible to diectly measue. Theoem Law of Sines[4] Whee a,b,c ae the sides of a spheical tiangle and A,B,C ae the angles opposing these sides, the Law of Sines states the following sin ( a sin(a = sin ( b sin(b = sin sin(c

15 Poof. Let thee be a spheical tiangle ABC, with adius. Let A be the vecto which goes fom the oigin of the sphee to the vetex A, B be the vecto which goes fom the oigin of the sphee to the vetex B, and C be the vecto which goes fom the oigin of the sphee to the vetex C Now conside x 1 = x 2 = C B 2 sin ( a A C 2 sin ( b x 3 = B A 2 sin ( c Each of these ae unit vectos and they also must be pependicula to the especting geat cicles, because of the ight hand ule. So x 1 is othogonal to the entie plane of secto COB. Then x 2 is othogonal to the entie plane coesponding to secto COA. Also x 3 is othogonal to the plane lying on secto BOA.

16 Conside the vecto identity below. ( A C ( C B = ( C ( B A C Fist note that, ( A C ( C B = ( ( b ( ( a 2 sin x 2 2 sin x 1 Thus, by we ae left with ( A C ( C B ( a ( b = 4 sin sin ( x 1 x 2 To detemine the value of ( x 1 x 2, fist conside that x 1 = x 2 = 1, since x 1 and x 2 ae unit vectos. Also note that C =. Next we need to conside the angle between x 1 and x 2, ecall that each of these angles is pependicula to its especting geat cicles.

17 Suppose we conside these unit vectos when they ae each lying exactly at point C. Because these unit vectos ae pependicula to the whole geat cicle we can abitaily place them at a specific point on that geat cicle without loss of geneality.we can abitaily place the unit vectos at a point because they do not depend on location, since they ae pependicula at any place on thei especting planes. Since x 1 is pependicula to the geat cicle coesponding to COB, then we know at vetex C, x 1 will be exactly tangent to the geat cicle ceated by vetices C and A. Note also that x 2 is pependicula to the geat cicle coesponding to AOC. Theefoe by simila logic x 2 will be exactly tangent at vetex C to the geat cicle which goes though vetices B and C. Theefoe these two vectos ceate an angle that must coespond to the angle ceated at vetex C. Thus ( x 1 x 2 = (1(1 sin(c C Substituting this value in ou pevious equation we get, ( A C ( C B ( a ( b = 3 sin sin sin(c C So theefoe by the oiginal vecto identity we have C ( B A ( a ( b = 3 sin sin sin(c Fom this the following is equivalent, A ( C B = 3 sin ( b sin sin(a B ( A C ( a = 3 sin sin sin(b

18 Also note that the tiple poduct of vectos coesponds to the deteminant. Fom popeites in linea algeba we know that deteminants do not change when two pais of ows ae swapped, which means it is invaiant. Theefoe, since we ae only swapping two pais of vectos; we know that the equations above all equivalent. ( a sin sin ( b sin(c = sin ( b ( a sin sin(a = sin sin sin(b Then dividing this whole expession by sin ( a sin ( b sin, we have which equivalently esults in sin(c sin ( = sin(a c sin ( = sin(b a sin ( b sin ( a sin(a = sin ( b sin(b = sin sin(c as desied. Now that we have the Pythagoean Theoem, Law of Cosines, and the Law of Sines fo a sphee, we can compae these to thei plana countepats. We can conside the following compaison. Spheical (Let =1 Pythagoean Theoem: Cosines: cos(c = cos(acos(b+sin(asin(bcos(c cos(c = cos(acos(b

19 Sines: sin(a sin(a = sin(b sin(b = sin(c sin(c Plana Pythagoean Theoem: c 2 = a 2 + b 2 Cosines: c 2 = a 2 + b 2 2abcos(C Sines: a sin(a = b sin(b = c sin(c Now it is inteesting to see that the Law of Sines is quite simila fo both plana and spheical. In contast the Pythagoean Theoem and Law of Cosines ae quite diffeent.it is also inteesting to note that in both plana and spheical the Pythagoean Theoem is a limit of the Law of Cosines. Fo the spheical equations this seems natual since we must use the Pythagoean Theoem in ode to pove the Law of Cosines; as shown in the poof above. It is impotant that we ecognize the fact that spheical tiangles and plana tiangles ae vey diffeent and the compaison of these equations help to show us the diffeences. Example Law of Sines Application Thee ae many examples we can use fo the application fo the Law of Sines. To keep vaiety I will including an example which elates to astonomy. Suppose that we have thee luna oves, call them A,B, and C espectively. Now suppose we wish

20 to find the distance between two of these such oves. We can use the following given infomation and the Law of Sines to discove a paticula length. Radius of moon:1737 km Distance between ove C and B (length a: 10 km Angle ceated at ove B: 3π 4 Angle ceated at ove A: π 6 How fa ae oves A and C (length of b? sin( a sin(a = sin( b sin(b Now we can substitute the numeical values and solve fo b so we have, b = 1737 acsin ( sin( 3π 10 sin( 4 sin( π b = km Thus the distance between ove A and ove C is aound 14 km. This value does not take into account any vaiation, such as dips o holes on the moon s suface so it is best used as a ough estimation. Again it was quite simple to find and gives some idea of how fa these luna oves ae fom eachothe. Theses ae just a few of the many examples spheical tigonomety has in ou wold. It is vey useful in the study of astonomy and in any navigational poblems on ou Eath. Having the ability to elate the spheical Phythagoeam Theoem, Law of Cosines, and Law of Sines to the adius of a sphee allows us use these equations on sphees of diffeent sizes. This also gives us moe feedom in the use of these equations as a whole, ove vaious disciplines such as math, physics, and astomony. Deiving the

21 spheical tigonometic fomulas in tems of the adius, was influential in histoy and its applications and uses continue to the moden day.

22 Bibliogaphy [1] Bummelen, Glen Van. Heavenly Mathematics: The Fogotten At of Spheical Tigonomety. Pinceton: Pinceton UP, Pint. [2] Clak, C. E. Poofs of the Fundamental Theoems of Spheical Tigonomety. Mathematics Magazine 21.3 (1948: 151. Web. [3] McCleay, John. Geomety fom a Diffeentiable Viewpoint. Cambidge: Cambidge UP, Pint. [4] Wilson, P. M. H. Cuved Spaces: Fom Classical Geometies to Elementay Diffeential Geomety. Cambidge: Cambidge UP, Pint. 22