Page images
PDF
EPUB

PROBLEM.

129. Given the radius of a circle equal to unity, to find the sine and cosine of l'. Fig. page 10.

Let AB be an arc of 60°, and let it be bisected in M, and MC drawn to the centre. Then the chord AB= radius (26) = 1, and AL = AB= radius, that is, radius being unity, the sine of 30°. Hence the cosine of 30°CL= ✓ CA2 — AL2=√1—=v=8660254. The cosine of 30° being found, the sine of 15° may be found as follows.

Let AB be the chord of an arc of 30°, then AB2 = 2 CM× ML (art. 47. cor. 1), therefore AB=2CM x ML, therefore AB=AL=√1 CM x ML. But ML •1339746, therefore the sine of 15°

CM-CL
✓CM× ML=5 x 1339746=2588190.

=

The sine of 15° being thus found, the sine of 7° 30' may be found in the same manner, and thence the sine of 3° 45', and so on, till after 12 bisections of the arc of 60° we find the sine of an arc of 52′′ 44′′ 3′′"" 45""""', which is 000255663462.

From the sine of an arc the cosine may be found, for ✔radius2 - sine2 cosine. Hence the cosine of 52" 44" 3" 45""" is found.

130. The sine of a very small arc is nearly equal to the length of the arc. For the number of the sides of an equilateral polygon inscribed in a circle may be so great, that the perimeter of the polygon and the circumference of the circle will differ by a line less than any given line (cor. 6. 1. sup.), or, which is the same thing, will be to each other nearly in the ratio of equality. Therefore their like parts will also be nearly in the ratio of equality; consequently the side of the polygon will be to the arc of the circle which it subtends nearly in the ratio of equality; therefore half the side of the polygon will be to half the arc which it subtends nearly in the ratio of equality, that is, the sine of any very small are will be to the arc itself nearly in the ratio of equality. Therefore if two arcs be very small, the first arc will be to the second very nearly as the sine of the first arc is to the sine of the second.

Hence the sine of an arc of 52" 44" 3""" 45"""" being known, the sine of an arc of 1' may be found; for 52′′ 44′′′′ 3′′'"' 45''''' : 1':: sine of the former arc 000255663462 : sine of the lat

ter. Thus, the sine of an arc of 1' is found to be ⚫0002908882. Hence the cosine of 1' is ⚫999999958.

131. The sine of 1' being thus found, the sine of 2', 3′, or of any number of minutes, may be found by the following proposition.

EC: EG :: DI : DM :: DB : DL (119)

or R cos. AC :: 2 s. BC: s. AD

-s. AB.

Let AB=a, BC=b, then AC=a+b, AD=a+2b. Hence R= 1: cos. (a+b) :: 2 s. b: s. (a+2b)—s. a, therefore s. (a+2b)—s. a=2 cos. (a+b) × s. b, therefore s. (a+2b)=2 cos. (a+b) x s. b+s. a. Now suppose b=1', a=0', 1', 2', 3', 4', &c.; then sine 2'2 cos. 1' x sine 1',

sine 3'2 cos. 2′ x sine 1' + sine 1',

sine 42 cos. 3' x sine 1'+sine 2'.

In general, sine n'=2 cos. (n-1)' x s. 1's. (n— 2)'. Hence, the sine and cosine of 1' being known, we obtain the sine of 2', and consequently the cosine of 2'. From the sine of 1' and the cosine of 2', we can find the sine and cosine of 3'. Thus may a table of the sines and cosines for every minute of the quadrant be computed; and as the multiplier, sine 1', is constant, the operation is not tedious.

When the arcs differ by more than 1' the method is the

same.

132. When you have found the sine and cosine for every minute of the arc to 30°, proceed as follows. By art. 121 sine (30°+B)=cos. B—sine (30°—B). Take B=1′, 2′, 3′, 4′, &c. then

sine 30° 1′ = cos. 1'-sine 29° 59′,
sine 30° 2′ = cos. 2′-sine 29° 58′,
sine 30° 3' cos. 3' ·sine 29° 57',

[blocks in formation]

Thus find all the sines, and thence all the cosines, as far as 45°. Then all the sines and cosines to 90° will be found; for the sine and cosine of an arc above 45° are respectively equal to the cosine and sine of an arc as much below 45°, one of these arcs being the complement of the other; that is, the cosine of 44° = sine of 46°, the cosine of 43° sine of 47°; also the sine of 44° cosine of 46°, the sine of 43° = cosine of 47°, &c.

=

[ocr errors]

133. The sines and cosines of all arcs to 90° being found, the versed sines are found by subtracting the cosines from radius (32).

134. The tangent of any arc is found by dividing the sine by the cosine of the same arc (41). When the tangents have been found in this manner as far as 45°, the tangents from 45° to 90° may be computed more conveniently, by another rule.

The tangent of an arc above 45° is the cotangent of an arc as much below 45°, and radius is a mean proportional between the tangent and the cotangent of any arc (41); therefore if D denote the difference between any arc and 45°,

tan. (459— D): 1 :: 1 : tan. (45° + D); therefore

[blocks in formation]

135. The tangents being found, the cotangents are known, for the tangent of an arc is the cotangent of the complement of that arc. Thus, the tan. 50° cot. 40°.

136. The radius is a mean proportional between the cosine and the secant of any arc, therefore if A be any arc,

[blocks in formation]

137. The secants being found, the cosecants are known, for the secant of an arc is the cosecant of its complement. Thus, the secant of 55° cosec. 35°.

138. When the sines have been found to every minute of the quadrant, the intermediate sines to every ten seconds of a degree, or even to every second, may be easily deduced by taking proportional differences. For the differences of the sines of arcs which are nearly equal will be as the differences of the arcs. Hence, if the sines of 39° 45' and 39° 46' be known, the sine of 39° 45′ 13′′ may be found as follows. Find the difference of the two given arcs, and also the difference of their sines; then 60" (the difference between 39° 45' and 39° 46') : 13" :: dif. of the sines of 39° 45′ and 39° 46': excess of the sine of 39° 45′ 13′′ above the sine of 39° 45'. Add the excess thus found to the sine of 39° 45', and you will obtain the sine of 39° 45′ 13".

This method will serve in all cases, when the arcs between which the interpolation is to be made are so great that the difference of 1' bears an inconsiderable proportion to either arc.

139. The natural sines, cosines, &c. being computed by the preceding rules, find their logarithms in the common tables of logarithms of the natural numbers, and add 10 to the indexes of their logarithms; the results will be the logarithms of the sines, cosines, &c.; and will be the same as those in the tables of logarithmic sines, tangents, &c. The logarithms in the

K

tables are supposed to be the logarithms of the sines, tangents, &c. to the radius 10000000000, which is the reason that 10 is to be added to the indexes of the logarithms of the natural numbers.

140. In the computation of tables the calculation should extend to two figures more than you intend to insert in the tables, that the last figure in the tables may have its nearest value, either greater or less than the true value. Thus, if the tables are to contain seven places of figures, you must compute nine figures; then the last two figures will always indicate the nearest figure to be inserted in the seventh place in the tables. If the computed value be 847962356, the nearest value to seven figures is 8479624. If the figure in the eighth place of the computed value be not 5, then eight figures will be sufficient; for if the eighth figure be greater or less than 5, the figure in the seventh place must be greater or less by unity than the value found in the calculation.

141. The old tables of logarithms contained the secants and versed sines, as well as the sines and tangents. But they are of so little use, and, when occasion for them occurs, are so easily derived from the other lines, that they are now omitted, and sines and tangents only are exhibited in the most improved modern tables.

Since the sine and cosine, the tangent and cotangent of any arc greater than a quadrant, are the same as the sine and cosine, the tangent and cotangent of its supplement, it is sufficient to exhibit the sines and tangents of arcs not exceeding the quadrant. If the canon give the sine or tangent of the arc AM, it also gives the sine or tangent of its supplement MBD. If I want to know the sine or tangent of 127° 15', I look for the sine or tangent of 52° 45'.

142. As many of the theorems or formulæ, which have been demonstrated in the first and third sections, are oftem employed in mathematical inquiries, it will be useful to collect them into a table, with some others which are deduced from them by simple processes.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors]

sine (a + b) sine (a —b)

2

[ocr errors]

cos.2 a =

Cos. a-cos. b = cos. (a + b) cos. (a - b)

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors]
« PreviousContinue »