47. To express the cosine of a side of a triangle in terms of sines and cosines of the angles. In the formula of Art. 37 we may, by Art. 28, change the sides into the supplements of the corresponding angles and the angle into the supplement of the corresponding side; thus cos (T - A)=cos (1-B) cos (T-C) + sin (1 – B) sin (7-C) cos(-a), 48. The formula in Art. 44 will of course remain true when the angles and sides are changed into the supplements of the corresponding sides and angles respectively; it will be found, however, that no new formulæ are thus obtained, but only the same formulæ over again. This consideration will furnish some assistance in retaining those formulæ accurately in the memory. 49. To express the sine, cosine, and tangent, of half a side of a triangle as functions of the angles. cos A + cos B cos C We have, by Art. 47, cos a = sin B sin C Let 2S = A +B+C; then B+C - A = 2 (S-A), therefore cos S cos (S – A) sin 2 sin B sin C a a and sin = { v cos Scos (S – A)) sin B sin C 2 a cos A + cos B cos C cos A + cos (B-C). Also 1 + cos a = =l+ i sin B sin c sin B sin c therefore cos } (A – B+C)cos } (A + B-C) cos (S-B) cos (S-C) cos 2 sin B sin c sin B sin c cos (S – B) cos (S – C) and 2 sin B sin c cos S cos (S – A) Hence tan cos (S – B) cos (S-C) The positive sign must be given to the radicals which occur in this Article, because is less than a right angle. , 2 . a COS -C}} a 50. The expressions in the preceding Article may also be obtained immediately from those given in Art. 45 by means of Art. 28. a a a COS and tan 2' 2 It 2 are real. For S is greater than one right angle and less than three right angles by Art. 32 ; therefore cos S is negative. And in the polar triangle any side is less than the sum of the other two; thus T-A is less than a B +7-C; therefore B+C - A is less than T ; therefore S – A is less than 5, and B+C – A is algebraically 2' greater than – T, so that S-A is algebraically greater than- ; 2 therefore cos (S – A) is positive. Similarly also cos (S – B) and cos (S – C) are positive. Hence the values of sin and tan 2' 2' are real, 51. Since sin a = 2 sin 2 2' we obtain T a a a COS a a COS sin a= 2 ca cos S cos (8 – A) cos (S – B) cos cos (S – B) cos (S-C)} – ?! We shall use N for {- cos S cos (S – A) cos (S – B) cos (S –C)}}. 52. To demonstrate Napier's Analogies. sin A sin B We have = m suppose ; sin 6 sin a Now cos A + cos B cos C = sin B sin C cos arm sin C sin b cos a, that is, and cos B+cos A cos C = sin A sin C cos b = m sin C sin a cos b, therefore, by addition, (cos A + cos B)(1 + cos C)=m sin C sin (a + b)......(3); therefore by (1) we have sin A + sin B sin a + sin 6 1 + cos C cos } (a – 6) c cot .(4). cos (a + b) 2 Similarly from (3) and (2) we have sin A - sin B sin 6 1 + cos C COS A + cos B sin (a + b) sin C sin } (a – 6) с that is, tan 1(A -B) cot (5). sin } (a+b) 2 By writing T - A for a, and so on in (4) and (5) we obtain tan 1 (a+b) = cos } (A – B) tan 2 cos (A + B) (6), sin 1 (A – B) tan 1 (a−b) = (7). sin 1 (A+B) The formula (4), (5), (6), (7) may be put in the form of proportions or analogies, and are called from their discoverer Napier's tan 2 sin a T с Analogies : the last two may be demonstrated without recurring to the polar triangle by starting with the formulæ in Art. 39. 53. In equation (4) of the preceding Article, cos 1 (a - b) and с cot 2 are necessarily positive quantities; hence the equation shews that tan (A + B) and cos } (a + b) are of the same sign; thus 1 (A + B) and 1 (a + b) are either both less than a right angle or both greater than a right angle. This is expressed by saying that 1 (A + B) and 1 (a + b) are of the same affection. a a 54. To demonstrate Delambre's Analogies. = {1+cos (a - b)} cos' 1C +{1+cos (a +b)} sino C; therefore cos} c = cos* } (a - b) cos' 1C + coso 1 (a + b) sino įC. Similarly, sin c=sin' 1 (a - b) cos® 1C + sino 1 (a+b) sino įC. Now add unity to the square of each member of Napier's first two analogies; hence by the formulæ just proved sinc + sec' (A – B) sin' } (a + b) sin 1C Extract the square roots; thus, since } (A + B) and } (a + b) are of the same affection, we obtain cos 1 (1+B) cos 1 c = cos } (a+b) sin 1 C .......(1), 2 Multiply the first two of Napier's analogies respectively by these results; thus sin } (A + B) cos 1 c = cos (a - b) cos C .......(3), ( sin ) (A – B) sin } c = sin } (a - b) cos } C .......(4). 1 – c The last four formulæ are commonly, but improperly, called Gauss's Theorems; they were first given by Delambre in the Connaissance des Tems for 1809, page 445. See the Philosophical Magazine for February, 1873. 55. The properties of supplemental triangles were proved geometrically in Art. 27, and by means of these properties the formulæ in Art. 47 were obtained; but these formulæ may be deduced analytically from those in Art. 39, and thus the whole subject may be made to depend on the formulæ of Art. 39. For from Art. 39 we obtain expressions for cos A, cos B, cos C'; and from these we find cos A + cos B cos C (cos a- cos b cos c) sin' a + (cos b - cos a cos c) (cos c- cos a cos 6) sino a sin b sin c In the numerator of this fraction write 1 - cos a fòr sin' a; thus the numerator will be found to reduce to cos a (1 – cos'a – cos' 6 – cosc + 2 cos a cos b cose), and this is equal to cos a sin B sin C sino a sin b sin c, (Art. 41); therefore cos A + cos B cos C = cos a sin B sin C. Similarly the other two corresponding formulæ may be proved. Thus the formulæ in Art. 47 are established; and therefore, without assuming the existence and properties of the Polar Triangle, we deduce the following theorem: If the sides and angles of a spherical triangle be changed respectively into the supplements of the corresponding angles and sides, the fundamental formula of Art. 39 hold good, and therefore also all results deducible from them. 56. The formulæ in the present Chapter may be applied to establish analytically various propositions respecting spherical triangles which either have been proved geometrically in the preceding Chapter, or may be so proved. Thus, for example, the second of Napier's analogies is sin ! (a - b) c tan } (A – B) = cot sin } (a+b) : |