PROPOSITION VIII. To find the expansion of a continuous function, such that its successive derivatives all become finite when the independent variable reduces to zero. Let y = fx be the function. It follows, from a process of reasoning precisely like that employed in the demonstration of the Binomial Theorem, that no other than integral additive powers of x can enter into the expansion; and it only remains (and is sufficient) to see if the assumption y = A2+ A1x + А ̧Ã2 + A‚Ã3 + ..., is possible, or, what amounts to the same thing, if the coëfficients Ao, A1, A2, A3, are determinable, and, therefore, real. ...9 Taking the successively derived functions, we find y' = A, 1+A,• 2x + A3 • 3x2 + ..., ່ : y" y"" = A, • 2 • 1 + A3 • 3 • 2x + A ̧ • 4 • 3x2 + ....., A,.3.2 · 1 + 4 • 4 • 3 • 2x + A5 • 5 •4• 3x2 + ..., A.4.3.2.1 + A; • 5 •. 4. 3. 2x + ..., Now, if in the above equations we make x = 0, and indicate the corresponding finite values of there results = A, yó=A, • 1, yő=A, • 2. 1, yo' = A, • 3 • 2 • 1, ...; This is essentially Maclaurin's Theorem, and is very serviceable in expansions, being more general than the Binomial, which it becomes simply by putting y = (a + x)". fx 10°. [la+bx + cx2+...)"]' = n(b + c • 2x + d • 3x2 + ...) [(298)] [(293)] a + bx + cx2 + dx3 + ... [(299)] ; [u=fx]. [[(298)]] BOOK SECOND. PLANE TRIGONOMETRY. SECTION FIRST. Trigonometrical Analysis. Construction. Describe the quadrant ABC; drop the perpendiculars BX, BY, upon the radii OA, OC ; produce OB so as to intercept AT and CV, perpendiculars drawn through the extremities of OA, OC, in T and V; then: Definition 1. The arcs AB, BC, are said to be complementary to each other. BC is the complement of X Fig. 52. AB and AB is the complement of BC; the arc 90° a is the complement of a and a is the complement of 90°-a; 45°+x and 45° are complementary arcs. Def. 2. The perpendicular BX is called the sine of the arc AB. Hence the sine of an arc is the perpendicular let fall from one extremity of the arc upon the diameter passing through the other extremity of the same arc. Def. 3. AT is the tangent of AB. Def. 5. BY (OX) is the sine of BC or the cosine of AB. Def. 8. AX is the verst sine of AB. Note. The abbreviations of the titles above, either with or without the period, are employed as symbols of the quantities themselves. Thus, if a denote any arc less than a quadrant and b any arc not greater than 45°, the above definitions give sin(45° + b) = cos(45° — b), cos(45°+b) = sin(45°—b). tana = cot(90° — a), cota = tan(90° — a); &c. (303) seca = cosec(90° — a), coseca = sec(90° - a); &c. (304) PROPOSITION I. The sum of the squares of the sine and cosine of an arc (305) is equal to the square of the radius, or to unity, when the radius is taken for the unit of the trigonometrical lines. Cor. The sine is an increasing and the cosine is a de- (306) creasing function of the arc, or the sine BX increases from 0 to r as the arc increases from 0 to 90°, while the cosine OX decreases from r to 0 for the same increase of the arc. See (194) and observe that the sine BX is half the chord of double the arc AB. PROPOSITION II. The tangent of an arc is to the radius as the sine to the (307) cosine; or, the tangent is equal to the sine divided by the cosine, if the radius be taken for unity. The radius is a mean proportional between the tangent (308) and the cotangent of an arc; or, the tangent and cotangent of an arc are reciprocals of each other when r= For, by similar triangles, we have AT = = 1. AO CO: CV, or tana: r=r: cota; tana cota = r2, = 1, when r = = 1. (fig. 52.) PROPOSITION IV. The square of the secant is equal to the sum of the (309) squares of the radius and the tangent. We have OT2 = OA2 + AT2, or sec2a = r2 + tan2a. (fig. 52.) The student may obtain other forms when wanted; as, for instance, the following: The secant is to the tangent as radius to the sine. Also (310) secant X cosine = r2 = 1. PROPOSITION V. AN INCREMENTAL VANISHING ARC is to be regarded as (311) a straight line perpendicular to the radius. Let AB be the arc in question; draw the tangents AT, BT, intersecting in T, and join OT; then will the triangles AOT, BOT, be equal, and OT will bisect the chord and arc in P and Q and be perpendicular to AB. From the similar triangles TAP, TOA, we have Fig. 53. B but QT = OT – OQ = (OA2 + AT2)* — OA, which reduces to [QT] = (OA2 — 02)1— OA = 0, when the arc AQB becomes = 0, since then AQ = ‡AQB = 0 and AT = tanAQ= 0; therefore the ultimate ratio of the vanishing quantities [AT], [AP], becomes But (113) the arc AQB is greater than the chord AB, and less AQB than the broken line ATB; .. the quotient is greater than AB |