LOGARITHMIC SINES, &C. log. sinxlog. - No. to {28596331+2log. x x 189 (378) As an example, let it be required to find the logarithm of the sine of 5°. Form (378) should not be employed when the arc exceeds 50°, and the last term, log. [1+ No. to (2log. x 1447)], may be omitted if the arc be less than 3°. Imitating the process above, we find log. cost No. to {13367543+2log. + log.[1+ - x = No. to (2log. - 0·778)]}. Operating as in the last example, we obtain (379) and log. sin5° 2.9402962, = log. cos5° = '9983442; log. tan5°29419520, log. cot5°1'0580480. In order to avoid minus characteristics, 10 is usually added; thus in most tables we find log. sin5° = 8‘9402960. Dividing the first of (320) by sina, there results sin(a+r) = cost+cosa sinx = cosr(1 + cota tang), .. log. sin(a + 2)=log. sina +log. cosa +M(cota tanz - cot2a tan2x +, −, ...). By a similar process we find log. cos(a + x) = log. cosa + log. cosã +tan'a tan2x +, −, ...). Let a = 5° and x = = 0°16', then +log. sina=2'9402962 log. M +log. cosz=1*9999993 + No. =0'0086639 ..log.sin5°1=2.9488741 Calculate the log. cos5°‘1. =T6377843 (1) log. cota 10580482) (2) = } log. tanz=32418778 ) (3) 3.9377103 4.2376363 2[(1)+(2)]+(1) 6·5375623 3[(1)+(2)]+(1) In order to avoid the accumulation of errors, the computations should be recommenced from new points of departure, for which purpose the above table of sines and cosines to every 3° may be employed. 2 (1 − x2)1 = (1 − x2)→, or (250), y' = (1)+ (− †) (1) ̄`−1 • (— x2)1 + ( − †) ( − + − 1) (1)→→→2, (− x2)2 where no constant is to be added, since x and y vanish together. (383) makes known z, then (x + z) is determined by (384) and finally x. For an example, let √216 • sinx +√72 • cosx = 12; we find x = I wx Cas 2 & RnZ Cod V 24 The second form will be preferable, when m is such a quantity as to be most readily computed by logarithms. What will the equations become when k=0? when l=0? The student may form an example for himself PROPOSITION XVIII. To resolve the quadratic equation, x2+2px = 9, when p and q are such as to require logarithmic tables. x + p = ±(p2 + q)* = ±(p2+p2 tan2v)* [putting q= p2 tan2v] =±p(1+ tan3v)* = ±p secv, Resolution of Triangles and Mensuration of Heights and Distances. PROPOSITION I. The length of a line and its inclination to a second line being given, to find its PROJECTION upon that line. Let a be the given line and l the line upon which its projection is to be made; drop the perpendiculars p, p, from the extremities of a upon l; then a, the portion of intercepted (a,1) cos (a,1) Fig. 58. between these perpendiculars, is called the projection of a upon l. Through the extremity of a nearest l, draw a parallel to a and terminating in P, then a 13. Also produce a to intersect l, PLANE TRIGONOMETRY. 193 making the angle (a,1); from the angular point and on the production of a, measure off the radius, r = 1, and from the extremdrop the perpendicular intercepting the cosine, cos(a,1); ity of The projection of a line is found by multiplying the (386) line into the cosine of its inclination to the line upon which it is projected. PROPOSITION II. To find an equation as simple as possible that shall embrace the relation existing between the sides and angles of a triangle. Let the sides of any triangle be denoted by a, b, c, and the angles respectively opposite by A, B, C. Then, dropping a perpendicular from C, we have (386) and a2 = acosB, b2 = bcos A; c=a2+ b2 = acosB+bcosA, i. e., C B A Fig. 59. Either side of a triangle is equal to the sum of the pro- (387) ducts formed by multiplying the two remaining sides into the cosines of their respective inclinations to the first mentioned line. This proposition, obviously little else than a corollary from (386), may be regarded as the fundamental theorem in the resolution of triangles; since it gives at once the equations, a = bcos C+ccos B, b = = acos C+ ccos A, C = acos B+bcos A ; } (3872) from which, by elimination, all possible relations among the sides and angles may be drawn. If one of the angles, as A, become greater than 90°, the corresponding side, b2 = bcos A, will be minus (338) .. c = a2 — b2 = acosB - bcosA, and the theorem still holds good. |