DIFFERENTIATION OF CIRCULAR FUNCTIONS. in which p and M'Q; the last, 445 are taken in their absolute value, the first part, p. Ax represents the line . Aa2, the line M'N, considering the two lines in an absolute sense. Remark. The supposition made at the beginning of this article, that positive abscisses go towards the right, and positive ordinates above the axis of the abscisses, will, as is usual, be adopted in all the figures, unless the contrary is specified. 26. We shall easily, then, with the assistance of the two preceding articles, find the differential of the arc of a curve. Suppose (fig. 1, 2, 3, 4) the arc EM; being a function of a, MN will be the difference of ; for the difference of x, that is to say PP' = Ax, and consequently MN Ap. The arc MN is greater than the chord MN, and less than the sum of the two lines MM'+ NM. Now, we have the chord MN = (MQ2+NQ2)3 = (25) [Ax2 + Ay2]3 = (5) [Ax2 + (p . Ax + ¥ . Ax2]3 = Ax (1 + p2 + 2μ¥. Ax + y2. Ax2)3. Developing the radical as a binomial, considering 1 + p2 as the first, and 2p¥ Ax+¥2. Ax2 as the second part, and using π to designate all the terms which include Ax, we find MM' = (MQ2 + M'Q2)3 = (25) (Ax2 + p2 . Ax2)3 = Ax (1 + p2)3, that is to say, the differential of the arc is equal to the square root of the sum of the squares of the differentials of the two variables. ON THE DIFFERENTIATION OF CIRCULAR FUNCTIONS. 27. Let us propose, for example, to find the differential of the arc of the circle. The equation of the circle referred to the centre is y2 + x2 = r2, whence dy -X = Whence results da y The differential of the arc of the circle enables us to find the differentials of all the trigonometrical lines by the following process. Supposing the radius of the circle = 1, we have for the equation of the circle referred to the centre, R B Let the arc BM = 4, the angle BCM will also be = , as the radius is equal to unity. It is again evident that AM = x = sin. 4 and AC = y = cos. p. Substituting these values of x and y in the differential of the arc of the circle d sin. dø = dx we obtain do = cos. or d. sin. cos. .dp: hence, the differential of the sine of an arc is equal to the cosine of the arc, multiplied by the differential of the arc. Now, we have already found d. sin. = cos. 4. d ; we have therefore cos. . d. cos. sin. =cos. o. do, that is, the differential of the cosine of an arc is equal to the sine of an arc, multiplied by the differential of the arc, and the result taken with the negative sign. sin. cos. = (art. 13) sin. 4. d. cos. Substituting the values previously found of d . sin. 4 and of d. cos. ø, we have cos. p2. do + sin. ¢2dø do that is, the differential of the tangent of an arc of a circle is equal to the differential of the arc, divided by the square of the cosine of the arc. that is, the differential of the cotangent of the arc of a circle is equal to the differential of the arc, divided by the square of the sine, the result being taken with the negative sign. By a similar process we find the differentials of other trigonometrical lines. * By the aid of the above formula, we may find the differentials of expressions involving sines, cosines, tangents, &c. We here present an example. [Let Remark.-According to this, we have also, if u represents a constant number, since d. (nx) =n. dx (art. 11), d . sin. nx = n . dx. cos. nx, d. cos. nx = - n. dx. sin. nx, 28. We may easily find from the differentials of the trigonometrical lines just obtained, the differentials of the arc in terms of the respective trigonometrical lines. Thus, we obtain from d. sin. = cos. 4 . dp, that is to say, the differential of the arc is equal to the differential of the sine of the arc, divided by the cosine of the arc. sin. . do, = that is, the differential of the arc is equal to the differential of the cosine, divided by the sine, the result taken with the negative sign. By a similar process we find from d. tan. p = do cos. $29 d. tan. sec. 82 1 + tan. ¿29 equal to the differential of the tangent, di that is, the differential of the arc is vided by the square of the secant. that is, the differential of the arc is equal to the differential of the cotangent, divided by the square of the cosecant, the result taken with the negative sign. By means of the formulæ in the preceding article the reader may easily express the differential of the arc by other trigonometrical lines. 29. We here subjoin for more convenient reference, the formulæ demonstrated above. Representing by a successively each of the trigonometrical lines of the arc 4, we have - = dø (1 — x2)ž› dø dø (1 — x2)3, dø = Let y(cos. x) sin x. Make cos. x = z, sin. x = u, then y = z′′, and dy = dz2 = 2a (du log. z +udz) [art. 32] [∞ (2 — x)]** In these formulæ the radius of the circle is (as in art. 27) supposed equal to unity. But it will be easy to refer them to radius = r, remembering that in the elements of Geometry, arcs as well as trigonometrical lines, relative to the same angles in circles of different radii, are as their respective radii. To exemplify this, let us designate a certain arc whose radius = 1 by p, and that whose radius = r by ør, and also the corresponding trigonometrical line by x and x,, we shall have : 4, = 1 : r and xx, = 1 : r, r Φ. whence = Substituting in place of x, dæ, o and do these values, in the above formulæ, we obtain the differentials of the arc and of the trigonometrical lines for the radius = r. Thus, for example, the formulæ for the sine Proceeding in this manner on all the above formulæ, we obtain for radius = r, Henceforward, however, in treating of trigonometrical functions, the radius will always be supposed 1, unless the contrary is expressly stated. = 30. We are now prepared to find the differential coefficient, and by it also the differentials of the superior orders of trigonometrical functions which contain arcs of the circle and trigonometrical lines: we can also, by art. 16, develope any trigonometrical function in a series, according to the integral powers of the independant. We have first in the function y sin. x (art. 27). dy da sin. x, r = cos. x, s = + sin. x, &c. ; =p= cos. x, q=— therefore (art. 16) yx_o= 0, Px=0 = 1, 9x=0 = 0, rx=0 = did =0 ·1, Sx_0 = 0, &c. x7 + 1 2 3 i. 2 3.4 5 1.2 3.4.5.6.7 + * whence, y = sin. x = x = وه sine of an arc by the arc itself. For y = cos. x; — cos. x, r = sin. x, s = cos. x, = 0, 9x=0 &c. 1, r_o= 0, $x=v = 1, &c.; x=0 1.2 1.2.3.4 1.2.3.4.5.6 a series which expresses the cosine of an arc by the arc itself. 31. By the same process by which we have expressed in a series the sine and cosine by the corresponding arc, we may also (art. 16) express every other trigonometrical line by its respective arc. But not dwelling on this, we propose to seek reciprocally a series for the arc expressed by its trigonometrical line. Thus, let y represent the arc whose sine, or according to the clearest notation, y arc (sin. = ∞) †; we have by art. 28 whence (art. 13) * There are many other methods of arriving at this valuable theorem, for one of which see the commencement of the Trigonometry, in vol. i. Dr. James Thomson deduces it from the following simple process: equating the coefficients of this with the corresponding ones of (a) in the assumed series (a), gives the series for sin. x in the text. In a similar manner, the series for cos. x would be found by assuming cos. x = 1+ Ax2 + Ba4+ &c. The student should recollect that the series for the sine and cosine possess the curious property of reproducing one another continually by repeated differentiations and divisions by da + See, however, the notes at pp. 41, 221, of this volume, as to a notation now brought much into use. |