Having thus computed the sine of a small arc that is nearly 1', the method of determining the sin. I' is, in principle, precisely the same as in Prob. 9. There is, however, a third method, considerably different in its principle, which finds the sine of 1' by the quinquesection and trisection of an arc. It may be thus explained, By the form [s], page 47, sin. 5 A = 5 sin. A 20 sin.3 A + 16 sin.5 A. Let 5 A = 30° then sin. 5 A= and A=6°; let 2 sin. A=x, then By approximation, find the value of x: thus, suppose a to be a near value, and a + v to be the true, then 1 = 5 (a + v) – 5 (a3 + 3a2 v) + as + 5 av, neglecting the terms that involve v2, v3, &c. consequently, 1-5a + 5 a3 - a5 Now, since sin. 5A = 2 of x, a = = .2: substitute in the expression for v this value, 10 and find the resulting value of v; it will appear to be = .009, the corrected value of a then, or a + v, is. 209; with this, find a new value of v, and another corrected value of a, and repeat the operation till x is found exact to a certain number of decimals, seven for instance; in which case x =.2090569, and consequently sin. A .1045285. Having thus obtained the sine of 6o, in the form [s""] page 47, that is, in sin. 3A = 3 sin. A 4 sin.3 A, put 3 A = 6o, and 2 sin. A=x, then the equation becomes .2090569=3x-x3. Find, as before, by the method and formula of approximation, a value of x, which, to seven places of figures, will be .0697989, consequently x or 2 sin. 2° = .0697989, and sin. 2° = .0346995. In order to find sin. 1o, take the form (p. 42,) sin. 2A=2 sin. A. cos. A, then sin. 2 A=4. sin. A − 4.sin. A: substitute for sin. 2A, or sin. 2o, its value, and by the solution of a quadratic equation find sin. A, and then, sin. A, or sin. 1o, the value of which to seven places of figures is 0174524. Repeat the operation, and 1° we have sin., or sin. 30', the value of which is .0087265. 2 By this method then we have descended from the sin. 30° to the sin. 1o and sin. 30'; and consequently, by like operations, we can descend from sin. 30' to sin. I' and sin. 30": and by this method, which is however extremely operose, we are able to find the sin. 1' without a proportion, and accordingly to avoid the use of a principle, which some may think doubtful; namely, that the sines of small arcs are to one another as the arcs themselves. The above method is, in fact, the same as that which is given with all its detail, at page 451, &c. in the sixth Volume of the Scriptores Logarithmici, edited by Baron Maseres. It is plain, however, that there is no necessity for beginning the computation from an arc of 30°; we may make it begin from any arc, the sine of which is known: for instance, by the form, page 67, sin. 9° = √(3 + √5) - √(5-√5) = .156434; since, therefore, sin. 9o = sin. (3.3°) = 3 sin. 3° 4 sin.3 3o, solve, as before, by approximation, the equation and the result gives x the sin. 3°. Again, solve a similar equation by the same mode and formula, and the result gives sin. 1o. And many like methods will suggest themselves to the mind of the intelligent Student. We shall now proceed to the second part of the construction of Trigonometrical Tables, the object of which will be understood from the succeeding Problem. PROBLEM 10. It is required from the sin. 30" and sin. 1', to compute the sines of 2, 3, 4, &c. minutes, and also the sines of 1, 2, 3, &c. degrees. By the form [a], page 31, sin. (A + B) = 2 sin. A. cos. B - sin. (A - B) = 2 sin. A (1-2 sin.) - sin. (A - B) 2 Let B=1' and let A successively equal 1', 2', 3', &c. then and thus may the sines of all succeeding arcs be computed, by a process not very tedious, since the only part of it at all long is the multiplication of sin. 2', sin. 3', &c. by the constant factor (2 sin. 30"), which is the square of the chord of 1'. In the above form substitute, instead of B, 1o, and, instead of A, successively 1, 2, 3, &c. degrees; then sin. 2o = sin. 1o + (sin. 1o sin. 3o = sin. 2° + (sin. 20 sin. 0) - 4 sin. 1o (sin. 30′′)2 and so on for the sines of all succeeding arcs. In order to compute the sines of arcs composed of degrees and minutes; arcs, for instance, such as 3o 2', 3o 3', substitute for B, 1', and for A successively 3o 1', 3° 2', 3° 3', &c. then sin. 3° 2' = sin. 3° 1' + (s'n. 3° 1' - sin. 3°) - 4 sin. 3° 1' (sin. 30") sin. 3° 3′ = sin. 3° 2' + (sin. 3° 2' - sin. 3° 1') - 4 sin. 3° 2' (sin. 30")2 &c. or, if we wish to compute for every ten minutes, put B = 10', and for A write successively A + 10', A + 20', A + 30', &c. thus, if A = 7°, sin. 7° 20′ = sin. 7° 10+(sin. 7° 10 - sin. 7°) - 4 sin. 7° 10'. (sin. 5')2 sin. 7° 30′ = sin. 7° 20'+(sin. 7° 20′ - sin. 7° 10')-4 sin. 7° 20′. (sin. 5'), &c. By the preceding methods we are enabled regularly to compute the sines of all arcs from 1' or 1" up to 90°; but, when the arcs exceed 60°, the application of the Trigonometrical formula, page 31, renders the arithmetical computation more simple and concise: thus, since sin. (B + A) = sin. (B - A) + 2. cos. B. sin. A. I Let B = 60°, then cos. B = cos. 60° =, consequently, - sin. (60° + A) = sin. A + sin. (60° - A). (see form 15 of Table, p. 39.) Hence, instead of the preceding, we may use this latter method, and compute the sines of all arcs exceeding 60°, by the simple addition of the sines of arcs previously computed: for instance, sin. 63° 5' = sin. 3° 5′ + 56° 55′ and, since sin. 3° 5' = .0537883, and sin. 56° 55′ = .8378775 the sin. 63° 5′ = .891665. The sine of all arcs from 0 to 90° being computed, the cosines of all the arcs of the quadrant are known; since cos. A = sin. (90° - A); for instance, cos. 63° 15' 7" = sin. 26° 44′ 53′′ cos. 13° 47′ = sin. 76° 13', &c. The sines and cosines being computed, the tangents may be computed from this expression, tan. A = sin. A and the co When the tangents of arcs up to 45o have been computed, the Trigonometrical formulæ previously given may be conveniently used in computing the tangents of arcs that lie between 45o and 90°: thus, by Prob. 4, page 35, tan. (A+B) = tan. A + tan. B = [if B = 45°] hence, as instances, putting A=1o, 2o, 3o, &c. 1 + tan. 1° tan. (45° + 1o), or, tan. 46°= tan. (45° +2°), or, tan. 47° = &c. &c. tan. (45° + A) = 1 + tan. 2° or, we may thus avoid the fractional form, 1 + tan. A 1- tan. A and tan. (45° – A) = ; 1-tan. A 1 + tan. A (1 + tan. A)-(1-tan. A) tan.2 A ... tan. (45° + A) - tan. (45° − A) = = 1 4. tan. A but, by the form of page 62, tan. 2 A = 2 tan. A • Hence, tan. (45° + A) = 2 tan. 2 A + tan. (45° - A), con sequently, tan. (45° + 1o), or, tan. 46° = 2 tan. 2° + tan. 44°, By these formulæ and methods may the sines, tangents, &c of arcs be computed. If we attend, however, to the history of the construction of Trigonometrical Tables, we shall find that all Tables have not been computed exactly by the same formulæ and methods: modern Tables, from the improved state of analytic science, have been computed by the most certain and expeditious methods. In the immense Tables du Cadastre, formed at the expense of the French Government, the sines of arcs are computed regularly by successive addition, according to the formulæ given in page 70; but, in such a construction, an error committed in the sine of an inferior arc would, it is plain, entail errors on |