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Given the radius CD=1, and DE the sine of 30°, to find DM, the sine of 15°.


DM=!DB=√(DE2+EB2) = {√ (·25+·017949193445) =.258819 sin 15°.


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 15°258819, and sin 30°, to find cos 15°.

258819 5 ·5(=ļrad) : ·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



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

S√(1— })=√ }={√3=•8660254=cos 30°.

By Prop. 2.




::·9659259: 4829629 FM 1: 8660254:: 258819 : 2241438=OP


7071067=FI=sin 45°.


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

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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 3°, 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 17o—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° 317320508; 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.

FIG. 4.


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


with CE or CB, which, when the arch DB is very small, will therefore be very nearly equal to each other; consequently the sine ED will also be very nearly equal to the tangent BG.

To illustrate this farther, draw DF parallel to CB; the triangles GDF, GCB are equiangular; therefore DF: CB:: GF: GB; whence if DF or EB is less than 10,000,000th part of the radius CB, GF, the difference between the sine and tangent will also be less than the 10,000,00th part of the tangent GB.

COR. Since an arch is greater than its sine, and less than its tangent, and the sine and tangent of a very small arch are nearly equal, the arch which lies between them must be nearly equal to either of them. Wherefore in very small arches, as

arch is to arch so is sine to sine.


To find the sine of an arch of one minute.

The sine of 30° being (equal to half the chord of 60°, that is, equal to half the radius or) equal to the sine of 15° may be found by Prop. 3. In like manner, from the sine of 15° we may obtain the sine of its half 7° 30', and so on by continually bisecting the arcs, and finding the sines, till after eleven bisections of the arc of 30°, we at last get the sine of 52" 54"" 3""" 45"""=000,255,663,462; the cosine of which viz. 999,999,967,318, is very nearly equal to the radius; in which case, as is plain from Prop. 7, the sines are proportional to the arcs. Therefore, as the arch 52" 44" 3"" 45""" is to an arch of 1', so is the sine of the former arch='000,255,763,462 to the sine of 1'000,290,888,204. When the sine of l'has been thus found, the sine and cosine of 2' may be found, by Prop. 4. and 2, to be 000,581,776,1 and 999,999,830,8 respectively.


If any angle (BAC) at the circumference of a circle be bisected by a straight line (AD), and AC one of the sides of the given angle be produced till it be cut by DE, which is made equal to AD, then will CE, the part produced, be equal to the chord AB.

In the quadrilateral figure ABDC inscribed in the circle, the angles B, and ACD, are together equal to two right

angles, (22. 3,) and therefore equal to the two angles ACD, DCE, (13. 1,) consequently B is=DCE. But (5. 1.) the angle E is DAC, because by construction DE is DA, and by the supposition DAC is DAB; therefore E=DAB. And the arches CD, DB and (29.3.) their chords are equal. Therefore (26. 1.) the triangles DAB, DCE are in every respect equal, and CE=BA.


If there be any number of equal arches, the chord of the first will be to the chord of the second, as the chord of the second is to the sum of the chords of the first and third; or as any one of the chords is to the sum of the preceding and following chords.

Let the arches AB, BC, CD, DE, EF, &c. be equal, and let the chords AB, AC, AD, AE, &c. be drawn, then will AB: AC:: AC: AB+AD::AD: AC+AE::AE: AD+ AF:: AF: AE+AG.

For produce AD to H, AE to I, AF to K, AG to L, so that each of the triangles ABC, ACH, ADI, AEK, AFL may be an isosceles triangle; then since the arch BC is equal to the arch CD, the angle BAD is bisected by AC, therefore (by Prop. 9.) DH=AB.

For the same reason, EI= AC; FK=AD; and GL-AE. And because the angles at A, being at the circumference, and standing on equal arches AB, BC, CD, DE, &c. are equal, the isosceles triangles ABC, ACH, ADI, AEK, AFL, are equiangular, and therefore similar to one another, (4.6.); whence AB: AC:: AC:(AH=)AB+AD: : AD: (AÌ=)AC+AE ::AE:(AK=)AD+AF::AF:(AL=)AË+AG; and so


COR. Since (by Prop. 4. Cor. 1.) rad : 2 cos AB:: chord AB: chord AC, (fig. 6,) therefore, rad: cos AB::AB: AC :: AC: AB+AD:: AD: AC+AE, &c., or (halving these chords,) rad 2 cos AB::AB: AC::AC: AB+ AD:: AD: AC+AE, &c.

Now, let each of the arches AB, BC, CD, &c. be 2'; then will AB be the sine of 1', AC the sine of 2', AD the sine of 3', AE the sine of 4', and so on.

Whence, having found the sine of 1' by Prop. 8, and its cosine, viz. 999,999,957,692, by Prop. 2, the sine of 2' is found by this analogy, rad: 2cos 1':: sin l': sin 2'; and from these,

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