Discepancies Between Euclidean and Spheical Tigonomety David Eigen
1 Non-Euclidean geomety is geomety that is not based on the postulates of Euclidean geomety. The five postulates of Euclidean geomety ae: 1. Two points detemine one line segment. 2. A line segment can be extended infinitely. 3. A cente and adius detemine a cicle. 4. All ight angles ae conguent. 5. Given a line and a point not on the line, thee exists exactly one line containing the given point paallel to the given line. The fifth postulate is sometimes called the paallel postulate. It detemines the cuvatue of the geomety s space. If thee is one line paallel to the given line (like in Euclidean geomety), it has no cuvatue. If thee ae at least two lines paallel to the given line, it has a negative cuvatue. If thee ae no lines paallel to the given line, it has a positive cuvatue. The most impotant non-euclidean geometies ae hypebolic geomety and spheical geomety. Hypebolic geomety is the geomety on a hypebolic suface. A hypebolic suface has a negative cuvatue. Thus, the fifth postulate of hypebolic geomety is that thee ae at least two lines paallel to the given line though the given point.
2 Spheical geomety is the geomety on the suface of a sphee. The five postulates of spheical geomety ae: 1. Two points detemine one line segment, unless the points ae antipodal (the endpoints of a diamete of the sphee), in which case they detemine an infinite numbe of line segments. 2. A line segment can be extended until its length equals the cicumfeence of the sphee. 3. A cente and a adius with length less than o equal to π whee is the adius of the sphee detemine a cicle. 4. All ight angles ae conguent. 5. Given a line and a point not on the line, thee ae no lines though the given point paallel to the given line. A line segment on a sphee is the shotest distance between two points on the sphee. This distance is the smalle section of ac fomed by the two points on a geat cicle. A geat cicle is a cicle whose cente is the cente of the sphee and whose adius is the adius of the sphee. In spheical geomety, a tiangle is the section of a sphee bounded by the acs of thee geat cicles. Thee ae two sections fomed in this way. One is the section whose inteio angles each have a measue less than 180. The othe is the section whose inteio angles each have a
3 measue geate than 180. I will be woking only with the fist tiangle, and will egad it as the spheical tiangle fomed by the acs of thee geat cicles. Also, each side must have a length less than o equal to half the cicumfeence of the sphee because othewise the side would not fit the definition of a line segment. I will focus on the following poblem: Fo a tiangle ABC whee a, b, and c ae the lengths of the sides opposite angles A, B, and C, espectively, given a, b, and the measue of angle C, what is the elationship between c whee ABC is in a Euclidean plane and c whee ABC is on the suface of a sphee, and what is the elationship between the aea of ABC in a Euclidean plane and ABC on the suface of a sphee? In othe wods, find c Euclidean c spheical and Aea Euclidean Aea spheical. The values of these atios will vay depending on the size and shape of the tiangles, as well as how lage they ae elative to the sphee. I will exploe these atios fo a vaiety of cicumstances. To do this, I will use spheical tigonomety. Because ABC is not necessaily ight, I will use tigonomety fo oblique tiangles. The key fomulae fo solving this poblem ae the law of cosines fo sides and the law of sines. The deivations of them ae much like the deivations fo the
4 law of cosines and law of sines in Euclidean tigonomety. The fist step is to deive a tigonomety fo ight tiangles. The measue of the angle fomed by the intesection of two geat cicles is defined to be equal to the measue of the angle fomed by the two lines tangent to each geat cicle at the point of intesection. It is also equal to the measue of the dihedal angle fomed by the planes of the geat cicles. (The measue of a dihedal angle is the measue of the angle fomed by the two lines of intesection made by the two planes foming the dihedal angle with a plane pependicula to the two planes line of intesection.) This is shown in the following poof. Call the point of intesection of the two geat cicles A and the cente of the sphee O. Let lines l and m each be tangent at A to one of the geat cicles that intesect at A. l OA and m OA because l and m ae each tangent to a cicle that OA is a adius of (and a line tangent to a cicle is pependicula to the adius of the cicle that intesects it). Because of this, the measue of the dihedal angle fomed by planes l-o and m-o is equal to the measue of the angle fomed by l and m, and theefoe equal to the measue of the spheical angle fomed by the intesection of the geat cicles.
5 1 In the pictue above, O is the cente of the sphee. ABC is a ight spheical tiangle with the ight angle at C. Constuct plane DEF though any point E on OB such that plane DEF OA. Let D be on OA, and F be on OC. DE OA and DF OA because DEF OA and DE and DF ae on DEF. Then ODF is a ight tiangle with its ight angle at D because DF OA, and ODE is a ight tiangle with its ight angle at D because DE OA. DEF OAC because DEF OA and OA is on OAC. If two planes intesect and ae each pependicula to a thid plane then thei line of intesection is pependicula to the thid plane; so EF OAC because BCO OCA and DEF OCA. OC is on OCA, theefoe EF OC, so OFE is ight. Also, DF is on OCA, theefoe EF DF, so DFE is ight. Fo each pat of spheical tiangle ABC, thee is an angle whose measue equals the measue of the pat (I will be expessing all angle and
6 ac measuements in adians). Each pat and coesponding angle ae: m A = m EDF m B = m DEF m C = m DFE a = m FOE b = m DOF c = m DOE Because of this, sin a = sin m FOE = FE OE = FE ED ED OE = sin A sin c. Using simila logic and constucting a plane pependicula to OB instead of OA, six moe fomulas ae obtained fo a total of seven. Using these seven fomulas an additional thee fomulas can be deived to make a total of ten fomulas. These ten fomulas ae: sin a = sin A sin c tan a = sin b tan A tan a = tan c cos B cos c = cos a cos b sin b = sin c sin B tan b = sin a tan B tan b = tan c cos A cos c = cot A cot B cos A = cos a cos B cos B = cos b sin A 2 To deive the law of sines and the law of cosines fo sides, the constuction of an altitude must be made. Constuct the altitude fom vetex C to side c and extend if necessay. The two cases (not extending side c and extending side c) a ae depicted hee:
7 3 Case I: c does not need to be extended. Case II: c does need to be extended. Law of Sines The identities of ight spheical tiangles listed befoe yield: sin h = sin a sin B, and sin h = sin b sin A Substituting fo sin h, sin a sin B = sin b sin A. Dividing both sides by sin A sin B, sin a sin A = sin b sin B. If the altitude h wee constucted fom vetex B to side b, the deived identity would be: By tansitivity, sin a sin A = sin c sin C sin a sin A = sin b sin B = sin c sin C. 4 Law of Cosines fo Sides In both cases, ADC is ight with D being the ight angle. Applying
8 identities fo ight spheical tiangles, cos a = cos h cos (c-m) fo the fist case, and cos a = cos h cos (m-c) fo the second case. Because cos (c-m) = cos (m-c), cos h cos (c-m) = cos h cos (m-c). Theefoe, the fist case can be used to deive the fomula fo both cases. Because cos (α-β) = cos α cos β + sin α sin β, cos h cos (c-m) = cos h (cos c cos m + sin c sin m) Applying the identities fo ight spheical tiangles to ADC, cos b = cos h cos m (dividing both sides by cos h gives cos m = cos b cos h ), sin m = tan h cot A, and sin h = sin b sin A. Substituting fo cos m and sin m, cos h (cos c cos m + sin c sin m ) = cos h ä ã å cos c cos b cos h ë ì + sin c tan h cot A Distibuting, cos h ä å cos b cos c ã cos h ë ì + sin c tan h cot A = cos c cos b + sin c sin h cot A Substituting sin h = sin b sin A and using the fact that sin A cot A = cos A, cos c cos b + sin c sin h cot A = cos c cos b + sin c sin b cos A.
9 Theefoe, by tansitivity, cos a = cos c cos b + sin c sin b cos A. By constucting the altitude h to sides a and b instead of c, these thee foms of the law of cosines fo sides can be deived: cos a = cos c cos b + sin c sin b cos A cos b = cos a cos c + sin a sin c cos B cos c = cos b cos a + sin b sin a cos C 5 Fo an example, I will use the tiangle fomed on the Eath by New Yok, Moscow and the Noth Pole. The distance between New Yok and Moscow will be unknown. The Eath can be teated as a sphee with a adius of 3959mi. Such a sphee is called the teestial sphee. On the teestial sphee, the equato is the geat cicle that detemines a plane pependicula to the line dawn between the Noth Pole and the South Pole. A meidian is a geat cicle that intesects both poles. The pime meidian is the meidian that Geenwich, England intesects. In the following pictue, the Noth Pole is at point N and the cente of the sphee is at point O.
10 The longitude of a point P on the sphee is the measue of the ac intecepted on the equato by the meidian that P is on and the pime meidian. In the above pictue, it is the measue of ac EF (which equals m EOF). Longitude is positive when P is West of the pime meidian and negative fo when P is East of the pime meidian. The latitude of a point P on the sphee is the measue of the ac intecepted on the meidian that P is on by P and the equato. In the above pictue, it is the measue of ac PE (which equals m POE). Latitude is positive when P is Noth of the equato and negative when P is South of the equato. The colattitude of a point P is π/2 - the latitude of point P. This is the measue of the ac intecepted on the meidian P is on by P and the Noth Pole (the measue of ac PN).
11 In the pictue above, N is New Yok, M is Moscow, and P is the Noth Pole. The distance between the Noth Pole and New Yok is New Yok s colattitude (NP), and the distance between the Noth Pole and Moscow is Moscow s colattitude (MP). The diffeence in longitude between New Yok and Moscow is the ac intecepted on the equato by the meidians they ae on. In the above pictue, this is ac EF. The measue of ac EF = m EOF. Because the plane the equato is in is pependicula to the line detemined by the Noth Pole and South Pole, m EOF equals the measue of the dihedal angle fomed by plane NEO and plane MFO (namely, N-PO- M). Because m EOF equals the measue of dihedal angle N-PO-M and the measue of N-PO-M is the measue of spheical angle NPM, m EOF equals the measue of spheical angle NPM. m EOF equals the measue of ac EF, which is the diffeence in longitude between New Yok and Moscow.
12 Theefoe, the measue of the angle fomed by New Yok and Moscow with Noth Pole (the Noth Pole as the vetex) is the diffeence in longitude between the two. New Yok s longitude is 37π/90, and its latitude is 2443π/10800. Moscow s longitude is -1127π/5400 and latitude is 223π/720. n = colattitude of N = π/2-2443π/10800 = 2957π/10800. m = colattitude of M = π/2-223π/720 = 137π/720. P = diffeence in longitude = 37π/90 + 1127π/5400 = 3347π/5400. By the law of cosines fo sides, cos NM = cos m cos n + sin m sin n cos P =.382398175, so NM = Accos.382398175 = 1.178405986 adians. To get the esult in miles, this is multiplied by the adius (3959mi) to get 4665.309297mi. If one used Euclidean tigonomety to calculate this, NM 2 = m 2 + n 2 2 mncos P = 23122014.49, so NM = 23122014.49 = 4808. 535587mi. The esult fom using Euclidean tigonomety is significantly lage than the esult fom spheical tigonomety. This is because the given lengths wap along the sphee, making the distance between them smalle than if they wee staight.
13 The poblem I pesented at the beginning of the pape is: Fo a tiangle ABC whee a, b, and c ae the lengths of the sides opposite angles A, B, and C, espectively, given a, b, and the measue of angle C, what is the elationship between c whee ABC is in a Euclidean plane, and c whee ABC is on the suface of a sphee and what is the elationship between the aea of ABC in a Euclidean plane and ABC on the suface of a sphee? In othe wods, find c Euclidean c spheical and Aea Euclidean Aea spheical. The tiangles in this poblem ae illustated hee: Euclidian Spheical The fist pat of the poblem is to find c Euclidean c spheical. The fist pat of finding this is to find the Euclidean c and spheical c independently.
14 Euclidean c 2 = a 2 + b 2 2 abcos C à c = a 2 + b 2 2 abcos C Spheical I will use the constant as the adius of the sphee. To apply law of cosines fo sides, eveything must be expessed in ac length. length 2 π = ac 2 π ac = length Applying the law of cosines fo sides, cos c = cos a cos b å cos a c = Accos ä ã cos b c = Accos ä å cos a ã cos b + sin a sin b cos C + sin a sin b cos C ë ì + sin a sin b cos C ë ì Theefoe, c Euclidean c spheical = a 2 + b 2 2 abcos C Accos ä å cos a ã cos b + sin a sin b cos C ë ì The second pat of the poblem is to find Aea Euclidean Aea spheical. The fist pat of
15 finding this is to find each aea independently. Euclidean Put ABC in a coodinate plane, with C at the oigin. sin C = y 1 1. à b sin C = y. Aea = ( )(base)(height), so Aea = b 2 2 ay. Substituting fo y, Aea = 1 2 absin C Spheical Using to law of sines to get A and B, sin A sin a = sin B sin b = sin C sin c. Theefoe, A = Acsin ä å sin a ã sin c ë sin C ì and B = Acsin ä å sin b ã sin c ë sin C ì The aea of a spheical tiangle is equal to 2 E, whee E is the angle excess of the tiangle. 6 The angle excess of a tiangle is the amount the
16 sum of the measues of its angles is ove π (the sum of the angles - π). Fo example, the 90-90 -90 tiangle (whose angles ae all 90, o π/2 adians) has an aea 1/8 of the suface aea of the sphee. Using the angle excess fomula, its aea = 2 2. The suface aea of a sphee is 4π 2, so the esult fom the angle excess fomula is tue. In the oiginal poblem, the angle excess fomula gives: Aea = 2 áa + B + C é. Substituting fo A and B, ä å Aea = 2 Acsin ã ä å sin a ã sin c ë sin C ì + Acsin ä å sin b ã sin c ë sin C ì + C ì ë Theefoe, Aea Euclidean Aea spheical = 2 ä å ã Acsin ä å sin a ã sin c 1 2 absin C ë sin C ì + Acsin ä å sin b ã sin c ë sin C ì + C ë ì I will exploe these atios fo the special case whee a=b=k on the unit sphee (whee the adius,, equals one). Because the only tiangles I am dealing with ae those whose angles each have a measue less than π
17 and geate than 0, 0 < C < π. Also, 0 < k < π because othewise k would not fit the definition of a line segment. Below ae gaphs of the c Euclidean c spheical atio as a function of C ( c Euclidean c spheical is on the y-axis and C is on the x-axis). Thee ae two gaphs: one whee k=π/2 and anothe whee k=1.2. k = π/2 y=1 k=1. y=1 As k inceases, c Euclidean c spheical inceases (no matte what the value of C is).
18 This implies that the spheical c becomes inceasingly smalle than the Euclidean c as moe of the sphee is coveed by sides a and b. The limit of c Euclidean c spheical as k appoaches zeo is 1. This implies that fo small values of k, the diffeence between Euclidean c and spheical c is negligible. An example of this is when one deals with local tiangles on the Eath. If one is dealing with tiangles that ae small elative to the size of the Eath (these ae about all of the tiangles encounteed in evey-day life), one can use Euclidean tigonomety and get esults just about as accuate as one would have gotten had they used spheical tigonomety. The limit of c Euclidean c spheical as C appoaches π is 1. This is because the length of side c is getting close to a+b (which always has the same value in the Euclidean tiangle as in the spheical tiangle). When k > π/2, thee is a liftoff of the entie gaph. The gaph of c Euclidean c spheical whee k=2.4 (a numbe geate than π/2) is pesented below:
19 k = 2.4 y=1 This liftoff is due to a wap-aound effect. Wap-aound occus when k is geate than one quate of the cicumfeence of the sphee. When this happens, the length of side c deceases as k inceases because sides a and and b ae wapping aound the place whee c is maximized. Thus, the maximum value fo c is when k=π/2. My poof of this is pesented hee: c = Accos á cos 2 k + sin 2 k cos C é Diffeentiating with espect to k yields: d c d k = ä å 1 ã 1 ácos 2 k + sin 2 k cosc é 2 ë ì á 2 cos k sin k + 2 sin k cos k cos C é Factoing out 2 cos k sin k,
20 d c d k = 1 1 ácos 2 k + sin 2 k cos Cé 2 á2 cos k sin k ácos C 1 éé d c d k = 2 cos k sin k ácos C 1 é 1 ácos 2 k + sin 2 k cos C é 2 The citical points ae at the values of k that make the deivative eithe equal to zeo o undefined. The deivative equals zeo when its numeato equals zeo, and it is undefined when the denominato equals zeo. Thus, the values of k that ae citical points ae values that make the numeato equal to zeo o make the denominato equal to zeo. Setting the numeato equal to zeo, Case I: the numeato equals zeo 2 cos k sin k ácos C 1 é = 0 Both cos C - 1 and -2 ae constants, so both sides of the equation can be divided by 2 ácos C 1 é without losing a oot. Executing this division, cos k sin k = 0 If a poduct of two numbes equals zeo, than one o both must equal zeo. Theefoe, cos k = 0 o sin k = 0 à k = π o k = 0 o k = π 2
21 Case II: the denominato equals zeo Setting the denominato equal to zeo, 1 ácos 2 k + sin 2 k cos C é 2 = 0 Squaing both sides, adding ácos 2 k + sin 2 k cos C é 2 to both sides, and then taking the squae oot of both sides, Substituting 1 cos 2 k fo sin 2 k, cos 2 k + á1 cos 2 k écos C = " 1 Distibuting cos C, cos 2 k + cos C cos 2 k cos C = " 1 Factoing out cos 2 k, á1 cos C écos 2 k + cos C = " 1 cos 2 k + sin 2 k cos C = " 1 Subtacting cos C fom both sides and dividing both sides by (1 - cos C), cos 2 k = " 1 cos C 1 cos C à k = Accos ä å " 1 cos C " ã 1 cos C ë ì Because the adicand must be positive and cos C < 1, the -1 of the " 1 must be discaded. Theefoe,
22 k = Accos ä å 1 cos C " ã 1 cos C ë ì à k = Accos á" 1 é = Accos á" 1 é à k = 0 o k = π The set of all the citical points is the union of the points fom Case I (numeato = 0) and the points fom Case II (denominato = 0). Theefoe, the citical points ae: k = π 2, k = 0, and k = π d c d k To find which ae elative maxima, I will make a sign diagam of d c d k. is epinted hee: d c d k = 2 cos k sin k ácos C 1 é 1 ácos 2 k + sin 2 k cos C é 2 (cos C - 1) is always negative because cos C < 1. The denominato is always geate than o equal to zeo because it is a positive squae oot.- 2 is always negative, so the sign of d c d k sign of cos k sin k. when k is not a citical point is the When 0 < k < π d c, cos k and sin k ae both positive, so 2 d k is positive.
23 When π 2 < k < π, cos k is negative and sin k is positive, so d c d k is negative when π 2 < k < π. The sign diagam is dawn below: ++++++++++ ----------- d c 0 π 2 π d k Thus, c is inceasing ove the inteval (0, π/2) and deceasing ove the inteval (π/2, π). Theefoe, thee is a elative maximum at k = π 2. Within the domain thee is no othe elative maximum, so k = π 2 is the absolute maximum fo c. Below ae two gaphs of Aea Euclidean Aea spheical as a function of C. Aea Euclidean Aea spheical is on the y-axis, and C is on the x-axis. In one gaph, k=1. In the othe, k=π/2.
24 k=1 y=1 k=π/2 y=1 The limit as k appoaches zeo of Aea Euclidean Aea spheical equals one. As k inceases, the y-intecept inceases. Also as k inceases, the limit as C appoaches π of Aea Euclidean Aea spheical deceases, and eaches a minimum of zeo at k = π/2. When k > π/2, the wap-aound effect ceates a liftoff of the gaph. A gaph whee k > π/2 is pesented hee:
25 k=2 y=1 In the amount of time I was given to complete this pape, I was unable to accomplish seveal things. I could not exploe the Aea Euclidean Aea spheical atio fo isosceles tiangles as much as I wanted. Also, I did not have enough time to exploe the two atios fo non-isosceles tiangles. I believe wapaound affects the atios fo non-isosceles tiangles when the sum a+b is geate than π/2 (π/2 on the unit sphee). In addition, I was unable to exploe the atios when the adius of the sphee is a constant. Howeve, the wok I did whee the adius equals one can apply to all
26 sphees; the units of length can always be defined such that the length of the adius of the sphee equals one. Fo example, on Eath one can measue eveything in tems of Eath adii.
27 NOTES 1. Pictue fom Paul R. Ride, Plane and Spheical Tigonomety, New Yok: The Macmillan Company, p. 201 2. Ten fomulas fom Ride, Plane and Spheical Tigonomety p. 203 3. Both pictues fom Ride, Plane and Spheical Tigonomety p. 212 4. Deivation adapted fom Ride, Plane and Spheical Tigonomety pp. 211-212. 5. Deivation adapted fom Ride, Plane and Spheical Tigonomety pp. 212-213 6. Angle excess fomula adapted fom Ride, Plane and Spheical Tigonomety p. 200