2x be found, and also y (=1+x); and from thence the required sides of the similar figure ABC, will, by proportion, be likewise known. PROBLEM XXXIII. Let there be three equi-different arches, AB, AC, and AD; and, supposing the sine and co-sine of the mean AC, of the lesser extreme AB, and of the common difference BC (or CD) to be given, it is proposed to find the sine and cosine of the greater extreme AD. C. n D Upon the radius AO let fall the perpendiculars Bb, Cc, and Dd; join B, D, and from the centre 0, let the radius OC be drawn, cutting BD in n also draw nR parallel to Cc, meeting AO in R: then, because of the similar triangles OCc and OnR, it will be, OC: On: Cc: nR; and, OC: On:: Oc: OR: whence we have B Ab CR d and OR = Ocx On : but, since BC is equal to CD (and therefore Bn equal to Dn), nR will, it is plain, be an arithmetical mean between Bb and Dd, and so is radius OC be supposed unity, will become Dd=Cc × 20n -Bb; and Od-Oc × 20n-Ob: from whence we have the two following theorems. 299 Theor. 1. If the sine of the mean of any three equi-different arches (the radius being supposed unity) be multiplied by twice the co-sine of the common difference, and from the product, the sine of either extreme be subtracted, the remainder will be the sine of the other extreme. Theor. 2. And if the co-sine of the mean of three equi-different arches be multiplied by twice the co-sine of the common difference, and the co-sine of either extreme be subtracted from the product, the remainder will be the co-sine of the other extreme. PROBLEM XXXIV. The sine and co-sine of an arch being given, to find the sine and co-sine of any multiple of that arch. Let the given arch be represented by A, its sine by x, and co-sine by y, the radius being unity. Then, since the arch A may be considered as an arithmetical mean between 0 and 2A, we shall, by the first of the two preceding theorems, have = sine of Axy. sine of 2A sine of 2A xy. sine of 3A sine of 4A = sine of SA x y = - xy) = xy3 -2xy; sine of 0) - sine of A) = sine of 2A= = xy; 3A sine of 5A (= sine of 4A x y-sine of SA= xy — Qxy3 · xy2 + x) = xy — Sxy2 + x; sine of 6A (= sine of 5A xy-sine of 4A = xys— Sxy3 yn n-6 1—6 × ya—7, &c. Moreover, from the second theorem, 3 we have co-sine of 2A (= co-sine of Axy-co-sine of 0= co-sine of 4a (= co-sine of 3A x y co-sine of 2A = 2 whence, universally, the co-sine of the multiple arch nA yn nyn-2 n n-3 will be truly represented by 2 + 2 2 -7 xy-3, &c. which series, as well as that for the sine, is to be continued till the indices of y become nothing, or negative. But, if you would have the sine expressed in terms of x only, then, because the square of the sine the square of the co-sine is always equal to the square of the radius, and therefore, in this case, x+4y-1, it is manifest that the sines of all the odd multiples of the given arch A, wherein only the even powers of y enter, may be exhibited in terms of a only, without surd quantities: so that 4-4x2 being substituted for its equal y2, in the sines of the aforementioned arches, we shall have 4th. Sine of 9A = 9x120x3 + 432x5 ·576x + 256x9; And, generally, if the multiple arch be denoted by nA, then the sine thereof will be truly represented by From this series the sine of the sub-multiple of any arch, where the number of parts is odd, may also be found, supposing (s) the sine of the whole arch to be given for let a be the required sine of the sub-multiple, and n the number of equal parts into which the whole arch is divided; then, by what has been already shown, we shall N n2-1 n2-9 have nx n n2-1 x5 &c.s: from the solution of which equation the value of x will be known. Hence also, we have an equation for finding the side of a regular polygon inscribed in a circle: for, seeing the sine of any arch is equal to half the chord of double that arch, let v and w be wrote above for x and s respectively, and then our equation will ben2-1 43 n2-1 no -9 X come no n 2. .3 X + X 2.3 4.5 -9 75 expressing the relation of chords, whose corresponding arches are in the ratio of 1 to n. But, when the greater of the two arches becomes equal to the whole periphery, its chord (w) will be nothing, and then the equation, by dividing the whole by n2-1 v2 n2-1 = 0; where n is the number of sides, and the side of the polygon. From the foregoing series, that given by Sir Isaac Newton, in Phil. Trans. mentioned in p. 242 of this Treatise, may also be easily derived. For, if the arch A and its sine x be taken indefinitely small, they will be to one another in the ratio of equality, indefinitely near, by what has been proved at p. 246; in which case, the ge neral expression, by writing A instead of x, will become n n2-1 n -1 nA n2-9 2.3 4.5 4.5 Therefore, if n be now supposed indefinitely great, so that the multiple arch nA may be equal to any given arch, the squares of the odd numbers, 1, 3, 5, &c. in the factors n-1, n2-9, n2-25, &c. may be rejected as nothing, or inconsiderable, in respect of n2; and then the foregoing series will become nA - n3A3 n'A7 2.3.4.5.6.7 + 2.3 n5A5 2.3.4.5 &c. wherein, if for nA, its equal ≈, and the same with that before given. Moreover, the aforegoing general expressions may be applied, with advantage, in the solution of cubic, and certain other higher equations, included in this form, |