- = Now, from the sine of 30°, those of 6o, 2o, and 1o, may easily be found. For, if 5A 30°, we shall have, from equation x, sin 5A = 5 sin a 20 sin3 A + 16 sin3 A: or, if sin A, this will become 16.rs 20x35x='5. This equation solved by any of the approximating rules for such equations, will give r = 1045285, which is the sine of 6°. 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 2°, 3x 4x31045285. This cubic solved, gives a0348995 sin 2o. Then, if s = sin 1o, we shall, from the second of the equations marked x, have 2s/1-s2 = 0348995; whence s is found0174524 sin 1°. Had the expression for the sines of bisected arcs been applied successively from sin 15°, to sin 7°30°, sin 3°45′, sin 1°524′, sin 564, &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 3°, and so on, a different series still would have been found. But 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. Ex. 2. Given the sine of 1°, to find the sine of 2°, and then the sines of 3°, 4o, 5o, 6o°, 7°, 8o, 9o, and 10o, each by a single proportion. Here, taking first the expression for the sine of a double arc, equá. x, we have sin 2°2 sin 1°√1 — sin2 1°='0348995. Then it follows from the rule in equa. xx, that : sin 1o sin 2°- sin 1° :: sin 2o + sin 1° : sin 3° = ·0523360 sin 2o: sin 3o-sin 1° :: sin 3° + sin 1° : sin 4° = '0697565 sin 3° sin 4-sin 1°:: sin 4o + sin 1°: sin 5° : =0871557 sin 4° : sin 5° — sin 1° :: sin 5o + sin 1° : sin 6° = ·1045285 sin 5°: sin 6o- sin 1o :: sin 6o + sin 1° : sin 7° 1218693 sin 6° sin 7° -sin 1° :: sin 7° + sin 1°: sin 8° =1391731 sin 7° sin 8°-sin 1° :: sin 8° + sin 1°: sin 9° sin 8° sin 9° - sin 1° :: sin 9° + sin 1° sin10° 1736482 To check and verify operations like these, the proportions should be changed at certain stages. Thus, : : =1564375 sin 1° sin 3° - şin 2° :: sin 3° + sin 2° : sin 5°, sin 1o : sin 4o — sin 3° : : sin 4° + sin 3o : sin 7o, sin 4° sin 7°— sin 3° :: sin 7° + sin 3o : sin 10°. The coincidence of the results of these operations with the analogous results in the preceding, will manifestly establish the correctness of both. Cor. Cor. By the same method, knowing the sines of 5o, 10o, and 15, the sines of 20°, 25°, 35°, 55°, 65°, &c, may be found, each by a single proportion. And the sines of 1o, 9°, and 10°, will lead to those of 19°, 29°, 39o, &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. 3. 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+B) = sin 45°, sin (n+1)B= sin 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"+1). Ex. 4. Explain the method of finding the logarithmic sines, cosines, tangents, secants, &c, the natural sines, cosines, &c, 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 101; 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 101×s= sine of the same arc or angle to rad 10. And this, in logs is, log 100s 10 log 10+ log s = 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 COS = Theref. log tan log rad + log sin-log cos 10+ log sin - log cos. rad2 tan COS rad2 Cosec= -- sin Therf. log cot2 lograd-log tan=20-log tan. Therf. log sec=2 log rad―log cos=20—log cos. Therf. l.cosec=2 log rad-log sin=20-log sin. Versed sine = = 2 x sin2 arc rad Therefore, log vers sin = log 2+2 log sin are 10. Er. 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 B and C = 3·8765577, 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 Atan B + tan c5·3047057 Take tan Atan B Remains tan c Take tan B + tan c =21445069=tan 65°. From tan Atan B + tan c = 5·3047057 Remains tan ▲ =3.8765577 = 14281480=tan 55°. Consequently, the three angles are 55°, 60°, and 65°. Er. 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. 111 and XXXIV, we have 2 sin Ass-a). (is—b). and sin c = √ (s-a). (s-b) ab Let the three sides of the required triangle be represented by x, x + 1, and r+2; the angle a being supposed opposite to the side r, and c opposite to the side +2: then the preceding expressions will become (+1). (x+2) 3x + 3 Assuming these two expressions equal to each other, as they ought to be, by the question; there results, after a little reduction, x3- 2x2 — 11 x − 2 — 0, a cubic equation, with one positive integer root = 4. Hence 4, 5, and 6, are the x sides of the triangle. 6 • 4 şin B = 7; sin c = 7; sincv7. The angles are, A = 41°409603 = 41° 24′ 84" 34", B = 55° 771191 55 46 16 18, 82.819206 = 82 49 9 8. Any Any direct solution to this curious problem, except by means of the analytical formulæ employed above, would be exceedingly tedious and operose. Er. 7. Demonstrate that sin 18° (− 1 + √5), and sin 54° = cos 36° is = cos 72° is = R R(1+√5). Er. 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 2, or 3, respectively. Ex. 10. Demonstrate that the sum of the squares 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. Er. 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? Er. 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. Er. 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°. Er. 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. CHAPTER IV. 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. Def. 3. A right-angled spherical triangle has one right angle the sides about the right angle are called legs; the side opposite to the right angle is called the hypothenuse. Def. 4. A quadrantal spherical triangle has one side equal to 90° or a quarter of a great circle. Def. 5. Two arcs or angles, when compared together, are said to be alike, or of the same affection, when both are less than 90°, or both are greater than 90°. But when one is greater and the other less than 90°, they are said to be unlike, or of different affections. ART. 2. The small circles of the sphere do not fall under consideration in Spherical Trigonometry; but such only as have the same centre with the sphere itself. And hence it is that |