(D) Hence, if A=any arc, and r-radius, then Asine A+ sine3 A 3 sines A 3.5'sine A 3.5.7⚫sine'A + + 2.4.5'4 2.4.6·7. 2-4-6-8-9 2.3.4.5.6.7.8°9′r3 The same series will apply to the chord of any arc by substituting the diameter d, or 2r instead of r*. (E) If 90°- A be substituted for A, then we shall obtain sine3 (902-A), 3 sine5 (90°-A) 90°-A sine (90°-A)+ + ,&c.; 2.3.2 2.4.5.g 3.5 cos? A &c.; or, + + 2.3.2 2.4.5 2.4.6.7 Cos Ar- +: 2r 2·3·4.r3 2·3·4·5·6·rs + &c. 2•3•4•5•6•7•8•p7) its chord, the radius being 1, is the 24th part of the cube of the length of the 24' arc nearly. See note, Chap. IV. (G. 77.) + Emerson's Trigonometry, 2d edit. page 32, &c. (G) A= THE CONSTRUCTION OF A TABLE OF SINES, &c. (I) There are various methods of constructing a table of sines, but the following, though not the least laborious, is the most simple. radius=cos 60° (K. 31.) Sine 30° These sines are found thus. From the square of the radius subtract the square of the sine, the square root of the difference is the cosine (M.98). The radius diminished by the cosine leaves the versed sine (R.99); and √vers2+sine2=chord*, (T. 99), the half of which is the sine of the next arc, &c. From the sine of 7°.30', find the sine of 3°.45', and so on continually, till the sines are as the arcs, which will be found at the end of the twelfth division from 30°: that is, at the end of the twelfth division, the arc will be 52′′.44".3".45", and its * In the formula here referred to, the minus should be plus. R 2 sine sine 0002556634; but at the end of the 11th division the arc is 1'.45".28' '.7"""'.30""", and its sine 0005113269; therefore the sine of the 12th division is just half the sine of the eleventh, in the same manner as the twelfth arc is half the eleventh. Now in indefinitely small arcs, the arcs will be to each other as their corresponding sines, hence 52". 44". 3.45".: 1':: ⚫0002556634, &c. : ·0002908882 the sine of one minute. The cosine of 'the square-root of the radius 1, diminished by the square of the sine of 1', viz. cosine l'-9999999577. And it is shewn (G. 116.) that 2 x cos. 1' x sine l'-sine 0'0005817764=sine 2′, or cos. 89°.58′ 2 x cos. I'x sine 2′- sine l'·0008726645=sine 3', or cos. 89 .57 2 x cos. l'x sine 3'-sine 2'-'0011635526=sine 4', or cos. 89.56 2 x cos. l'x sine 4'-sine 3'0014541406 sine 5', &c. &c. This method may likewise be extended to the cosines, by beginning at the other end of the arc, for the cosine is only the sine of the complemental arc. 2 x cos. I'x cos. l'—cos.0′9999998308cos. 2′, or sine89°.58′ 2 x cos. l'x cos. 2′-cos. 1'=9999996192—cos.3', or sine 89 .57 2 x cos. l'x cos. 3.3'-cos. .2′ =•9999993231=cos.4′, or sine 89 .56 2 x cos. l'x cos. 4' — cos. 3′ =*9999989423=cos.5', or sine 89 .55. Proceed thus to find the sine and cosine for every minute of the arc as far as 30°. (K) Then, if A and B be any two arcs, A 1 sine A.COS B rad sine (A+B) + sine (A−B) E. 109.; let A=30°, then sine Aradius (K. 31.); and cos Bsine (30°+B)+sine (30°-B); hence, sine (30°+B)=cos B-sine (30°-B). If, therefore, B=1'.2'.3'.4', &c. successively, we shall have Sine 30°.2′ = cos 2'-sine 29°.58′ Sine 30°.3' cos 3'-sine 29°.57, &c. And in this manner find all the sines, and thence all the cosines (M. 98.) as far as 45°. By these means all the sines and cosines from 0 to 90° will be obtained; for sine (45°+A)= cos (45°-A) and cos (45°+4)=sine (45°-A). (L) The sines and cosines being constructed, Cosine sine radius: tangent (U. 97); hence the tangents are found. : : Tangent radius:: radius: co-tangent (Z. 97); hence the cotangents are found. : Cosine radius:: radius: secant (X. 97); hence the secants are found. Sine radius: radius: co-secant (Y. 97); hence the co-secants are found. The versed sines are found by subtracting the cosine from radius, radius, or adding the cosine to the radius, if the arc be greater than a quadrant. (M) ARTIFICIAL, or LOGARITHMICAL, SINES, &c. are only the logarithms of the natural sines, &c. The natural sines are generally calculated to the radius 1, and being in all cases, except when the arc is 90°, less than radius or 1, they must of course be decimals. Now the logarithm of a whole number and the logarithm of a decimal is the same, only the index or whole number prefixed to the former is affirmative, and that to the latter negative. Hence if logarithmical sines, &c. were immediately formed from natural sines which are calculated to the radius 1, their indices would all be negative. To avoid this, the logarithmical radius instead of being taken 1, as in the natural sines, is generally considered as ten thousand millions. (N) To find the logarithmical sine of 1' to 7 places of figures, without the index. Rad. 1: rad. 10,000,000,000 :: 0002908882 (the natural sine of 1', the radius being 1):2908882 (the natural sine of 1', the radius being 10,000,000,000.) The logarithm of 2908882 6.4637261 the logarithmical sine of 1'. Hence the indices to the logarithmical sines from O' to 4', will be 6; from 4' to 35' the indices will be 7; from 35' to 5°.45′ the indices will be 8; from thence upwards to 89 they will be 9; and at 90° the index will be 10=the logarithm of the radius 10,000,000,000. (0) Hence if we take the natural sine of any arc from a table of natural sines (where the radius is unity), and multiply it by 10,000,000,000; the logarithm of the product will give the logarithm sine of that arc to as many places of figures as the natural sines are carried to. It may be proper to inform the learner, that this method will not be exactly true for the first five degrees; because the natural sines in the tables are not carried to a sufficient number of places. (P) In logarithms the operation of multiplication is performed by addition, and division by subtraction. The logarithm sines being constructed, the tangents, &c. are formed thus: Sine+10-cosine tangent (U. 97.) 20 Cosine = 20-sine =co-secant (Y. 97.) If you double the logarithmical sine of half an arc, and subtract 9.6989700 from the product; the remainder will be the logarithmical versed sine of that arc. = For radius versed sine square sine arc (I. 101.) or, radius x versed sine square sine arc, or, log. 5,000,000,000 +log. +log. versed sine 2x log. sine arc, or, 9.6989700 + log. versed sine 2 x log. sine arc. Therefore, log. versed sine =2x log. sine arc-9.6989700. (Q) The following logarithmical formulæ for the sine, cosine, &c. of any arc, may in some cases be useful, viz. Sine costang 10 tang+ 10 sec = cos + 10−cot = 20-cosec. Cossinę +10-tang = 20 sec sine+cot-10=cot + 10-cosec. Tang=sine+10-cos-20-cot. Cot cos+10-sine=20-tang. Sectang+10-sine 30-sine-cot=cosec+10-cot= 20- cos. Coseccot+10-cos 30-cos-tang sec +10-tang= 20-sine. SOLUTIONS OF THE DIFFERENT CASES OF RIGHT ANGLED PLANE TRIANGLES. The different cases both of right angled plane triangles, and oblique angled plane triangles, have already been solved by logarithms. The solutions here given are adapted to the table of natural sines, and the method of notation is that used by Lagrange and Legendre, the three angles of the triangle being represented by A, B, C, and their opposite sides by a, b, c, as in the figure annexed; r=radius=sine of 90°. CASE I. Given the angles and the hypothenuse, to find the base and perpendicular. SOLUTION. a= sine A.b cos c.b_tang A .b r.b C CASE II. Given the angles and the base, to find the hypothenuse and the perpendicular. CASE III. Given the angles and the perpendicular, to find the base and the hypothenuse. SOLUTION. |