Imágenes de páginas
[blocks in formation]

Other solutions of the last two cases are sometimes used. For example, Case 3. may be resolved by Prop. V, (fig. 18.)


BA: AC:: rad: tan ABC.

Rad: tan (ABC-45°) :: tan (B+C): tan (B-C.)

Now, since (B-C) is thus found, and (B+C) is given, B and C may be found by Lemma III. Then, sin C: sin A ::AB: BC.

And Case 4. may be resolved by Prop. X, (fig. 12. 13.) Thus:

BC: CA+AB:: CA-AB: F. Then F is either the difference or sum of CD, DB, the segments of the base, according as F is less or greater than BC. In either case, the sum of CD, DB, and their difference being given, CD and DB may be found by Lemma III. Then,

CA: CD::rad: cos C.

BA: BD:: rad: cos B. Thus C and B are found, and consequently A.



THE Trigonometrical Canon is a set of tables exhibiting the lengths of the sines, tangents and secants of every arch of a quadrant, containing an exact number of minutes, (or seconds,) in parts of the radius, the radius itself being 1: so that the sine, tangent and secant of any arch or angle may be found

directly from such tables; and conversely, any sine, tangent and secant being given, the arch or angle which it expresses may also be readily found.

The trigonometrical canon may be constructed by means of the following propositions:


Any two sides of a right angled triangle being given, the other side may be found.

For, since (47. 1.) AC2=AB2+BC2; therefore AC2— BC=AB, and AC-AB1=BC; and by extracting the square root, AC will be (AB+BC*); AB= √(AČ3— BC2); and BC= √(AC1—AB2).


The sine (DE) of an arch (DB) and the radius (CD) being given, to find the cosine (DF).

Since, in the right angled triangle CDE, the two sides CD and DE are given, the other side CE, (which is equal to DF the cosine,) will, by the preceding proposition, be = √(CD2 -DE2).


Given the radius CD=1, and DE, the sine of an arch of 30°, to find DF, the cosine of 30°.

DF=CE=√(CD2—DE2) = √(1—·25) = √ ·75=

•8660254cos 30o.


The sine (DE) of any arch (DB) being given, to find (DM or BM) the sine of half that arch.

Since DE and the radius DC are given, CE may be found, (by Prop. 2.) and also EB, which is the difference between the radius and the cosine CE.

Then, in the right angled triangle DBE, the two sides DE and EB being given, the other side DB may be found, (by Prop. 1.); the half of which DM or BM is the sine of (the arch DL or LB=) half the given arch BD.


Given the radius CD=1, and DE the sine of 30°, to find DM, the sine of 15°. EB-CB-CE=1-86602541339746.


[blocks in formation]


The sine (BM) of an arch (BL) being given, to find the sine of double that arch.

Find (by Prop. 2.) CM the cosine of the given arch BL; then, as radius is to cosine, so is twice the given sine to the sine sought.

For, the triangles CBM, DBE, having the angle at B common, and the angles at M and E right angles, are equiangular; therefore (4. 6.) CB: CM:: (DB=) 2BM : DE; that is, rad: cos: : 2 sin : DE.

COR. 1. Hence, CB: 2CM::DB: 2DE; that is, as radius to twice the cosine of half an arch, so is the chord of the whole arch to the chord of double that arch.

COR. 2. Hence also CB: 2CM:: (2BM: 2DE: :) BM : DE::CB: CM; that is, as the sine of a given arch is to the sine of double that arch, so is half the radius to the cosine of the given arch.


Given radius 1, sin 15o="258819, and sin 30°= 1⁄2, to find cos 15°.

258819: 5:5(=1rad) : ·9659259=cos 15°.


The sines (FO, DK) of any two arches (FD, BD) being given, to find (FI) the sine of their sum, and (EL) the sine of their difference.

Draw the radius CD, through the point O draw OP parallel to DK, produce FO till it meets the circumference in E, and draw through the points O, E, the straight lines OM, EG parallel to CB, and meeting FI in M and G.

Since the sines FO, DK are given, the cosines CO, CK may be found by Prop. 2; and because the triangles CDK, COP, CHI, FOH, FOM, are equiangular, the sides about

the equal angles are proportional, (4. 6.); therefore CD: DK :: CO: OP; which will thence be known: and CD: CK :: FO: FM; which will thence be also known.

But OP MI, (34. 1.); therefore OP+FM=MI+FM -FI the sine of FB the sum of the arches. And since OM is parallel to EG the base of the triangle FGE, as FO: OE :: FM: MG. But FO=OE, (3. 3.); therefore FM=MG =ON, (34. 1). Whence OP-FM-OP-ON NP EL the sine of EB the difference of the arches. Q. E. I.


COR. Since the differences of the arches BE, BD, BF are equal, the arch BD is an arithmetical mean between the arches BE and BF; or, the greater of two unequal quantities is an arithmetical mean between the sum and difference of those quantities.


Given radius 1, sin 30°=4, and sin 15°=258819, to find the sine of (30°+15°) 45°.

[blocks in formation]

If three arches are equidifferent, the radius is to double the cosine of the mean arch, as the sine of the common difference of the arches is to the difference of the sines of the extreme arches.

For, let BE, BD, BF be three equidifferent arches, their common difference is the arch FD, of which FO is the sine, and FG is the difference between FI and EL the sines of the extremes. It was shown in Prop. V. that CD: CK:: FO: FM; wherefore, (doubling the consequents,) CD: 2CK:: FO: (2FM) FG. Q. E. D.

COR. 1. When BD is an arch of 60°, FO the sine of the common difference of the arches is equal to the difference between FI and EL, the sines of the extreme arches.

For in that case CK, the cosine of BD, is the sine of 30°, the double of which is equal to the radius CD; consequently FO, FG, are equal.

COR. 2. Hence, when the sines of all arches, that are distant from one another by a given interval, from the begin

ning of the quadrant to 60°, are given, the other sines may be found by one addition only, viz. by adding the sine of the common difference to the sine of the less extreme. Thus, the sine of 61° is sin 59° + sin 1°; the sine of 62° is sin 58° +sin 2°; the sine of 63° is=sin 57° + sin 3o, and so on,

COR. 3. When the sines of all arches from the beginning of the quadrant to any part of it, that differ from one another by a given difference are given, we may thence find the sines of all the arches to the double of that part. For example, let all the sines from the beginning of the quadrant to 15° be given, then all the sines for every degree to 30°, may be found by the analogy in this proposition.

For rad: 2 cos 15°:: sin 1°: sin 16°-sin 14°. In like manner, rad: 2 cos 15°:: sin 2° : sin 17°-sin 13°; and rad: 2 cos 15°: : sin 3°: sin 18°—sin 12°; and so on. Whence, if the differences between the sines of 14° and 16°, of 13° and 17°, of 12° and 18°, be added to the sines of 14°, 13°, and 12° respectively, the sums will be the sines of 16°, 17° and 18° respectively.


In like manner, when all the sines to that of 30° have been found, rad 2 cos 30°:: sin 1°: sin 31°-sin 29°; and rad: 2 cos 30°:: sin 2° : sin 32° —sin 28°; and rad : 2 cos 30°: : sin 3°: sin 33°—sin 27°; and so to 60°. In this case, since cos 30° is√3, therefore 2 cos 30° = √3=1·7320508; by which if the sines of the common differences be multiplied, the products will be the differences of the sines of the extremes respectively.

By the same method, when the sine of 1' only is known, we may find all the sines from 1' to 15°; from 15° to 30°; and from 30° to 60°, to every minute of the quadrant. For, having the sin l', we may find the cos l' by Prop. 2; and the sin 2' by Prop. 4, and cos 2' by Prop. 2; then rad: 2 cos 2' :: sin l': sin 3'--sin l'; which added to sin l' gives sin 3′; and

so on.


In very small arches, the sine and tangent of the same arch are to each other nearly in a ratio of equality.

For, because the triangles CED, CBG are equiangular, CE: CB:: ED: BG. According as the point D approaches to B, EB will evidently become less and less in comparison

« AnteriorContinuar »