The arcs OT" and OT"" are in the third and fourth quadrants, respectively, and their co-secants, CS and CS', are estimated from the center and the termini of the arcs, and are therefore negative; hence, The co-secant of an arc in the third or fourth quadrant is negative. The word co-secant is an abbreviation of complementi secans, the secant of the complement. In fact, CS, the co-secant of OT, is the secant of OT, the complement of OT; hence, The co-secant of an arc is the secant of its complement. CR, the secant of OT, is the co-secant of O'T, the complement of OT; hence, The secant of an arc is the co-secant of its complement. Let the arcs OT and T'P be equal. Then, since T'P is the supplement of OT', OT will be the supplement of OT'. O'TOT', since they are complements of equal arcs. Hence, the angle OCT, measured by the arc O'T, is equal to the angle O'CT', measured by the equal arc O'T'. The right angles, COS and CO'S', are equal. Hence, in the triangles having the common side CO', and the two adjacent angles equal, CS is equal to CS'; but CS is the co-secant of OT, and positive, and CS' is the co-secant of OT', and positive; hence, The co-secant of an arc is equal to the co-secant of its supplement. The co-secant of 0' is +. As the arc increases from 0° to 90°, the co-secant decreases from ∞ to +1. As the arc increases from 90° to 180', the co-secant increases from +1 to +∞. As the arc increases from 180° to 270°, the co-secant passes through, changes its sign from to, and decreases numerically, but increases algebraically from - to 1. As the are increases from 270° to 360', the co-secant increases ∞ ――― numerically, but decreases algebraically from 1 to -∞. Hence, the limiting values of the co-secant are cosec 0°+∞, +1, cosec 180° +∞, cosec 270°-1, cosec 360°: ∞o. cosec 90° To aid the memory, and for convenience of reference, we give the following tabular summaries: Functions. sine. co-sine. = versed-sine. co-versed-sine. tangent. co-tangent. secant. co-secant. 0° 47. Signs of the Circular Functions. = = = 90° 1st q. ++++++++ = 180° = 48. Limiting Values of the Circular Functions. 2d q. 1 + + 1 + 3d q. +1 cvs cot 1 sec = 1 1 + + + +1 | 1 sin sin 0 sin +1 sin =+0 sin 0 +1 cos =+0 cos -1 cos 10 0 cos +1 vsin+0 vsin=+1 vsin=+2 vsin=+1 vsin=+0 = COS -= CVS tan +1 cvs = =+0 cvs +0 tantan cot =+∞ cot =+0 cot sec = +1 sec + ∞ sec cose∞ cose = +1 cose: 270° - = 4th 9. +∞ cose 1 | + | | +++ | 360° +2 cvs =+1 0 co tan ∞ sec 1 cose: ∞ +1 ∞ 49. Problem. To find any function of an angle to the radius R, in terms of the corresponding function of the same angle to the radius 1, and the reverse. Let sin C denote sin C to the radius CT-1, and sin CR denote sin C to the radius CT' R. From similar triangles, CT CT':: MT or 1 : R sin C sin CR. .. (1) sin CR=sin C1× R. .. (2) sin C1= Let formulas for other functions be deduced; hence, 1. Any function of an angle to the radius R is equal to the corresponding function of the same angle to the radius 1, multiplied by R. 2. Any function of an angle to the radius 1 is equal to the corresponding function of the same angle to the radius R, divided by R. sin CR R TABLE OF NATURAL FUNCTIONS. 50. Description of the Table. This table gives, to the radius 1, the values of the sine, co-sine, tangent, and co-tangent, to five decimal places, for every 10′ from 0° to 90°. For sines and tangents, the degrees are given in the left column, and the minutes at the top. For co-sines and co-tangents, the degrees are given in the right-hand column, and the minutes at the bottom. S. N. 4. 51. Problem. To find the natural sine, co-sine, tangent, or co-tangent of a given arc or angle. Let us find the natural sine of 35° 42′ 24′′. The difference between the natural sines of 35° 40′ and 35° 50′, as given in the table, is .00236. Now 2′ 24′′.24 of 10', which is found thus: 60|24 10 2.4 Then take Nat sin 35° 40'-.58307 .. Nat sin 35° 42′ 24′′.58364 In case of co-sine or co-tangent, the correction must be subtracted, since, between 0° and 90°, the greater the angle, the less the co-sine and co-tangent. 52. Examples. 1. Find the natural sine of 75° 45′ 30′′. .24 2. Find the natural co-sine of 15° 36′ 12′′. Ans. .96927. 3. Find the natural tangent of 43° 33′ 18". 53. Problem. Ans. .96315. Ans. .95079. 4. Find the natural co-tangent of 84° 28' 30". Ans. .09673. To find the angle corresponding to a given natural sine, co-sine, tangent, or co-tangent. 1. Find the angle corresponding to the natural sine .50754. Looking in the table we find the angle 30° 30'. 2. Find the angle whose natural sine Difference corresponding to 10' 3′ 21′′. ... Correction 10'X: 164 ... Angle = 55° 30′ 3′ 21′′ = .82468. = .82413. 54. Examples. 1. Find the angle whose sine is .75684. = 55° 33′ 21′′. In case of co-sine and co-tangent, the angular difference must be subtracted, since the greater the co-sine or co-tangent, the less the angle, for values between 0° and 90°. Ans. 49° 11′ 13′′. 2. Find the angle whose co-sine is .67898. Ans. 47° 14' 10". 3. Find the angle whose tangent is 1.34567. Ans. 53° 22′ 53′′. 4. Find the angle whose co-tangent is .98765. Ans. 45° 21′ 22′′. TABLE OF LOGARITHMIC FUNCTIONS. 55. Description of the Table. The table of logarithmic functions gives to the radius 10,000,000,000 the logarithm of the sine, co-sine, tangent, and co-tangent, for every minute, from 0° to 90°. The expression, logarithmic sine, tangent, etc., is equivalent to the logarithm of the sine, of the tangent, etc. For sines and tangents, the degrees are given at the top of the page, and the minutes in the left-hand column. |