Next, to find the sine of 2°, we have again, from equation x, sin 3A = 3 sin a 4 sin3 A: that is, if x be put for sin 2o, 3x 4x3 =·1045285. This cubic solved, gives * = 0348995 = sin 2. = Then, ifs sin 1o, we shall, from the second of the equa tions marked x, have 2s 1-s2 - 0348995; whence s is found = 0174524 = sin 1o. Had the expression for the sines, of bisected arcs been applied successively from sin 15°, to sin 7°30°, sin 3°45′, sin 1°52', sin 561, &c, a different series of values might have been obtained or, if we had proceeded from the quinquisection of 45°, to the trisection of 9°, the bisection of 3o, and so on, a different series still would have been found. what has been done above, is sufficient to illustrate this method. The next example will exhibit a very simple and compendious way of ascending from the sines of smaller to those of larger arcs. But Ex. 2. Given the sine of 19, to find the sine of 2o, and then the sines of 3o, 4o, 5o, 6o, 7o, 8o, 9°, and 10°, each by a single proportion. Here, taking first the expression for the sine of a double = arc, equa. x, we have sin 2° 2 sin 1°/1-sin2 1°='034895. Then it follows from the rule in equa. xx, that sin 1° sin 2°-sin 1° :: sin 2o + sin 1° : sin 3o 0523360 sin 2° : sin 3°-sin 1° :: sin 3° + sin 1° : sin 4° 0697565 sin 3°: sin 4o-sin 1o :: sin 4° + sin 1°: sin 5° = 0871557 sin 4° : sin 5o-sin 1o :: sin 5o + sin 1° : sin 6° =·1045285 sin 5o : sin 6° -- sin 1o :: sin 6° + sin 1° : sin 7o 1218693 sin 6° : sin 7°-sin 1o :: sin 7° + sin 1o: sin sin 7° : sin 8° -sin 1o :: sin 8o + sin 1° : sin 9° sin 8: sin 9°-sin 1° :: sin 9° + sin 1° : sin 10° To check and verify operations like these, the proportions should be changed at certain stages. Thus, 8° =1391731 •1564375 •1736482 sin 1° sin 3° - sin 2° :: sin 3° + sin 2° : sin 5°, sin 1o : sin 4o - sin 3o :: sin 4° + sin 3°: sin 7o, sin 4o : sin 7o — sin 3o :: sin 7° + sin 3°: sin 10o. The coincidence of the results of these operations with the analogous results in the preceding, will manifestly establish the correctness of both. Cor. By the same method, knowing the sines of 5o, 10o, and 15, the sines of 20°, 25°, 35o, 55°, 65°, &c, may be found, each by a single proportion. And the sines of 1o, 9o, and 109, will lead to those of 19°, 29°, 39°, &c. So that the sines may be computed to any arc and the tangents and other trigonometrical lines, by means of the expressions in art. 4, &c. Ex. S. Find the sum of all the natural sines to every minute in the quadrant, radius: 1. In this problem the actual addition of all the terms would be a most tiresome labour: but the solution by means of equation XXVII, is rendered very easy. Applying that theorem to the present case, we have sin (A + Inв) = sin 45°, sin (n+1)Bsin 45°0'30", and sin B-sin 30. Therefore sin 45° X sin 45° 0'30" = 3438 2467465 the same sum required. sin 30" From another method, the investigation of which is omitted here, it appears that the same sum is equal to (cot 30"+). Ex. 4. Explain the method of finding the logarithmic sines, cosines, tangents, secants, &c, the natural sines, cosines, &G, being known. The natural sines and cosines being computed to the radius unity, are all proper fractions, or quantities less than unity, so that their logarithms would be negative. To avoid this, the tables of logarithmic sines, cosines, &c, are computed to a radius of 10000000000, or 1010; in which case the logarithm of the radius is 10 times the log of 10, that is, it is 10. Hence, if s represent any sine to radius 1, then 1010 × 8 = sine of the same arc or angle to rad 1010. And this, in logs is, log 1010 s = 10 log 10+ log 8 = 10+ log s. The log cosines are found by the same process, since the cosines are the sines of the complements. The logarithmic expressions for the tangents, &c, are deduced thus: Tan rad sin Therf. log tan = log cos. cos = 10 + log sin rad2 rad2 Cot Sec= Cosec = Versed sine log rad + log sin-log Therf. log cot-2 log rad-log tan=20-log tan. Therf. log sec=2log rad-log cos=20-logcos. Therf.l.cosec=21ograd-logsin=20-logsin. Therefore, log vers sin log 2+2 log sin arc 10. Ex. 5. Given the sum of the natural tangents of the angles A and B of a plane triangle = 3.1601988, the sum of the tangents of the angles в and c = 31-8765577, and the continued product, tan A. tan B. tan c = 5.3047057 to find the angles A, B, and c. It has been demonstrated in art. 36, that when radius is unity, the product of the natural tangents of the three angles of a plane triangle is equal to their continued product. Hence the process is this: From tan A+ tan в + tan c =5.3047057 Take tan A+ tan B Remains tan c = 3 1601988 =2.1445069 tan 65°. = 5.3047057 3.8765577 1.4281480 tan 55°. Consequently, the three angles are 55°, 60°, and 65°. Ex. 6. There is a plane triangle, whose sides are three consecutive terms in the natural series of integer numbers, and whose largest angle is just double the smallest. Required the sides and angles of that triangle? If A, B, C, be three angles of a plane triangle, a, b, c, the sides respectively opposite to A, B, C ; and s = a + b + c. Then from equa, III and xxxiv, we have sin A = 2 — — vis (¦s—a). (§s—6) . (¿s—c). and sinc bc = ↓ (†s −a) . (¿s − b) ab Let the three sides of the required triangle be represented by x, x + 1, and x + 2; the angle A being supposed opposite to the side x, and c opposite to the side + 2: then the preceding expressions will become sin A = 2 (x+)(1+2) sin !c=✓ 3x+3 x+3x+1 x= Assuming these two expressions equal to each other, as they ought to be, by the question; there results, after a little reduction, x3 5x2 12x-2 O, a cubic equation, with one positive integer root x = 4. Hence 4, 5, and 6, are the sides of the triangle. sin A =√15.7.3.3 = 2222 3/8 √ 15.15.7 = 2:15 17 117. sin B =V7; sin c = 7; sinc = 2.3='VT. 6 ✓7; The angles are, a = 41°.409603 = 41°24′ 34" 34"", 6 Any direct solution to this curious problem, except by means of the analytical formulæ employed above, would be exceedingly tedious and operose. Solution Solution to the same by R. ADRAIN. D B A C Let ABC be the triangle, having the angle ABC double the angle A, produce AB to D, making BD = BC, and join cp; and the triangles CBD ACD are evidently isosceles and equiangular; therefore BD or BC is to CD or AC as AC to AD. Now let AB, BC = X- 1, AC = x + 1, then AD = 2x 1, and the preceding stating becomes x - 1: x + 1 : : x + 1 : 2x-1, which by multiplying extremes and means gives 2x2 3x+1= x2 + 2x + 1, and by subtraction x 5x, or dividing by x, simply x 5, hence the sides are 4, 5, 6. The same conclusion is also readily obtained without the use of algebra. Ex. 7. Demonstrate that sin 18° (− 1 + √5), and sin 54° cos 36° is = = cos 72° is fr (1 + √ 5). R Ex. 8. Demonstrate that the sum of the sines of two arcs which together make 60°, is equal to the sine of an arc which is greater than 60, by either of the two arcs: Ex. gr. sin 3+ sin 59° 57′ = sin 60° 3′; and thus that the tables may be continued by addition only. Ex. 9. Show the truth of the following proportion: As the sine of half the difference of two arcs, which together make 60°, or 90°, respectively, is to the difference of their sines; so is 1 to 1/2, or 3, respectively. Ex. 10. Demonstrate that the sum of the square of the sine and versed sine of an arc, is equal to the square of double the sine of half the arc. Ex. 11. Demonstrate that the sine of an arc is a mean proportional between half the radius and the versed sine of double the arc. Ex. 12. Show that the secant of an arc is equal to the sum of its tangent and the tangent of half its complement. Ex. 13. Prove that, in any plane triangle, the base is to the difference of the other two sides, as the sine of half the sum of the angles at the base, to the sine of half their difference also, that the base is to the sum of the other two sides, as the cosine of half the sum of the angles at the base, to the cosine of half their difference. Ex. 14. How must three trees, A, B, c, be planted, so that the angle at A may be double the angle at B, the angle at в double that at c; and so that a line of 400 yards may just go round them? Ex. 15. In a certain triangle, the sines of the three angles are as the numbers 17, 15, and 8, and the perimeter is 160. What are the sides and angles? Ex. 16. The logarithms of two sides of a triangle are 2-2407293 and 2-5378191, and the included angle, is 37° 20'. It is required to determine the other angles, without first finding any of the sides? Ex. 17. The sides of a triangle are to each other as the fractions, what are the angles? Ex. 18. Show that the secant of 60°, is double the tangent of 45°, and that the secant of 45° is a mean proportional between the tangent of 45° and the secant of 60°. Ex. 19. Demonstrate that 4 times the rectangle of the sines of two arcs, is equal to the difference of the squares of the chords of the sum and difference of those arcs. Ex. 20. Convert the equations marked xxxiv into their equivalent logarithmic expressions; and by means of them and equa. Iv, find the angles of a triangle whose sides are 5, 6, and 7. SPHERICAL TRIGONOMETRY, SECTION I. General Properties of Spherical Triangles. ART. 1. Def.1. Any portion of a spherical surface bounded by three arcs of great circles is called a Spherical Triangle. Def. 2. Spherical Trigonometry is the art of computing the measures of the sides and angles of spherical triangles. VOL. II. E Def |