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PROP. II. PROB.

Fig. 29.

Fig. 29.

The sine DE of the arc BD, and the radius CD, being

given, to find the cosine DF.
The radius CD, and the sine DE, being given in
the right angled triangle CDE, there will be given (by
the last Prop.) CDq-DEq=(CE=) DF.

PROP. III. PROB.
The sine DE of any arc DB being given, to find DM or

BM, the sine of half the arc.
DE being given, CE (by the last Prop.) will be
given, and accordingly EB, which is the difference be-
tween the cosine and radius. Therefore DE, EB,
being given, in the right angled triangle DBE, there
will be given DB, whose half DM is the sine of the
arc DL= the arc BD.

Fig. 29.

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PROP. IV. PROB.
The sine BM of the arc BL being given, to find the sine

of double that arc,
The sine BM being given, there will be given (by
Prop. 2.) the cosine CM. But the triangles CBM,
DBE, are equiangular, because the angles at E and M
are right angles, and the angle at B common : where-
fore (by 4. 6.) we have CB : CM::(BD, or) 2 BM :
DE. Whence, since the three first terms of this ana-
logy are given, the fourth also, which is the sine of the
arc DB, will be known.

Cor. Hence CB : 2 CM::BD: 2 DE; that is, the
radius is to double the cosine of one half of the arc DB,
as the subtense of the arc DB is to the subtense of
double that arc. Also CB : 2 CM::(2 BM : 2 DE::)
BM : DE:: CB : CM. Wherefore the sine of an
arc, and the sine of its double, being given, the cosine
of the arc itself is given.

PROP. V. PROB.
The sines of two arcs, BD, FD, being given, to find FI,

the sine of the sum, as likewise EL, sine of their dif-
ference.
Let the radius CD be drawn, and then CO is the

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Fig. 30.

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cosine of the arc FD, which accordingly is given, and draw OP through O parallel to DK; also let OM, GE, be drawn parallel to CB. Then because the triangles CDK, COP, CHI, FOH, FOM, are equiangular; in the first place, CD : DK::CO: OP, which, consequently, is known. Also we have CD : CK::FO : ÉM; and so, likewise, this will be known. But because FO=EO, then will FM=MG=ON; and so OP+FM-FI=sine of the sum of the arcs; and OP

- FM: that is, OP-ON=EL=sine of the difference of the arcs : which were to be found.

Cor. Because the differences of the arcs BE, BD, BF, are equal, the arc BD is an arithmetical mean between the arcs BE, BF.

PROP. VI. THEOR.

The same things being supposed, the radius is to double Fig. 50.

the cosine of the mean arc, as the sine of the difference is to the difference of the sines of the extremes.

For we have CD: CK :: FO: FM; whence by doubling the consequents, CD: 2 CK::FO:(9 FM, or) to FG, which is the difference of the sines ÈL, FI.

Q. E. D,

Cor. 1. If the arc BD be 60 degrees, the difference of the sines FI, EL, will be equal to the sine FO of the difference. For, in this case, CK is the sine of 30 degrees; the double whereof is equal to the radius (by 15. 4.); and so, since CD=2 CK, we shall have FO=FG. And, consequently, if the two arcs BE, BF, be equidistant from the arc of 60 degrees, the difference of the sines will be equal to the sine of the difference FD.

COR. 2. Hence, if the sines of all arcs distant from one another by a given interval, be given, from the beginning of a quadrant to 60 degrees, the other sines may be found by one addition only. For the sine of 61 degrees=the sine of 59 degrees + the sine of one degree; and the sine of 62 degrees=the sine of 58 degrees + the sine of two degrees. Also, the sine of 63 degrees=the sine of 57 degrees +the sine of 3 degrees, and so on.

Cor. 3. If the sines of all arcs, from the beginning of a quadrant to any part of a quadrant, distant from each other by a given interval, be given, thence we may find the sines of ail arcs to the double of that part: For

may be found.

example: let all the sines to 15 degrees be given ; then, by the preceding analogy, all the sines to 30 degrees

For the radius is to double the cosine of 15 degrees, as the sine of 1 degree is to the difference of the sines of 14 degrees, and 16 degrees; so, also, is the sine of s degrees to the difference between the sines of 12 and 18 degrees; and so on continually, until you come to the sine of so degrees.

After the same manner, as the radius is to double the cosine of 30 degrees, or to double the sine of 60 degrees, so is the sine of 1 degree to the difference of the sines of 29 and 31 degrees :: sine 2 degrees to the difference of the sines of 28 and 32 degrees :: sine 3 degrees to the difference of the sines of 27 and 33 degrees. But, in this case, the radius is to double the cosine of 30 degrees, as 1 to v 3.

For (see the figure for Prop. 15. Book IV. of the Elements) the angle BGC=60 degrees, as the arc BC, its measure, is a sixth part of the whole circumference; and the straight line BC=R. Hence it is evident that the sine of 30 degrees is equal to half the radius; and therefore, by Prop. 2. the cosine of 30 degrees= R2_V 3R?

and its double=3R’=R x V 3. 2 Consequently, radius is to double the cosine of 30° : : R:RX 3:11:3.

And, accordingly, if the sines of the distances from the arc of 30 degrees, be multiplied by 3, the differences of the sines will be had.

So, likewise, may the sines of the minutes in the beginning of the quadrant be found, by having the sines and cosines of one and two minutes, given. For, as the radius is to double the cosine of 2'::sine 1' ; difference of the sines of l' and 3'::sine 2' : difference of the sines of oʻ and 4'; that is, to the sine of 4'. And so, the sines of the first four minutes being given, we may thereby find the sines of the others to 8', and from thence to 16', and so on.

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Fig. 31.

PROP. VII. THEOR.
In small arcs, the sines and tangents of the same arcs are

nearly to one another, in a ratio of equality.
For, because the triangles CED, CBG, are equi-
angular, CE : CB:: ED : BG.. But as the point E

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approaches B, EB will vanish in respect of the arc BD; whence CE will become nearly equal to CB, and so ED will be also nearly equal to BG. If EB be less

1 than the

10000000 part of the radius, then the difference between the sine and the tangent will be also

1 less than the

part of the tangent. 10000000 Cor. Since any arc is less than the tangent, and greater than its sine, and the sine and tangent of a very small arc are nearly equal; it follows, that the arc will be nearly equal to its sine: and so in very small arcs, it will be, as arc is to arc, so is sine to sine.

PROP. VIII. PROB.

To find the sine of the arc of one minute.
The side of a hexagon inscribed in a circle, that is,
the subtense of 60 degrees, is equal to the radius (hy
Coroll

. 15th of the 4th); and so the half of the radius will be the sine of the arc of so degrees. Wherefore the sine of the arc of 30 degrees being given, the sine of the arc of 15 degrees may be found (by Prop. 3.). Also the sine of the arc of 15 degrees being given (by the same Prop.) we may have the sine of 7 degrees 30 minutes. So, likewise, can we find the sine of the half of this, viz. 3 degrees 45 minutes ; and so on, until 12 bisections being made, we come to an arc of 52, 44*, 03, 45', whose cosine is nearly equal to the radius; in which case (as is manifest from Prop. 7.) arcs are proportional to their sines : and so, as the arc of 52°, 44", 03*, 45*, is to an arc of one minute, so will the sine before found be to the sine of an arc of one minute, which therefore will be given. And when the sine of one minute is found, then (by Prop. 2. and 4.) the sine and cosine of two minutes will be had.

PROP. IX. THEOR.

If the angle BAC, being in the periphery of a circle, be Fig. 32.

bisected by the right line AD, and if AB be produced until DESAD meets it in E; then will СE=AB.

In the quadrilateral figure ABDC (by 22. 3.) the angles B and DCA are equal to two right angles= DCE+DCA (by 13. 1.): whence the angle B=DCE.

Fig. 33.

But, likewise, angle E=DAC (by 5. 1.) = DAB, and
DC=DB: wherefore the triangles BAD and CED
are congruous, and so CE is equal to AB.

Q. E. D.
PROP. X. THEOR.
Let the arcs AB, BC, CD, DE, EF, &c. be equal : and

let the subtenses of the arcs, AB, AC, AD, AE, &c.
be drawn ; then will AB: AC :: AC: AB+ AD: :
AD: AC + AE:: AE : AD+ AF:: AF: AE+AG.

Let AD be produced to H, AE to I, AF to K, and AG to L, so that the triangles ACH, ADI, AEK, AFL, be isosceles ones: then because the angle BAD is bisected, we shall have DH=AB (by the last Prop.): so likewise EI=AC, FK=AD, also GL=AE.

But the isosceles triangles ABC, ACH, ADI, AEK, AFL, because of the equal angles at the bases, are equiangular : wherefore it will be, as AB : AC:: AC: (AH=) AB + AD:: AD: (AIO) AC + AE:: AE: (AK=) AD + AF:: AF :(AL=) AE+AG. Q. E. D.

Cor. 1. Because AB is to AC, as radius is to double the cosine of the arc AB (by Coroll. Prop. 4.), it will also be, as radius is to double the cosine of the arc AB, so is } AB: 1 AC:: 1 AC: JAB+AD:: $ AD: 3 AC+ AE:: 1 AE: 1 AD+i AF, &c. Now let each of the arcs AB, BC, CD, &c. be 2'; then will # AB be the sine of one minute, i AC the sine of 2 minutes,

AD the sine of 3 minutes, 1 AE the sine of 4 minutes, &c. Whence, if the sines of one and two minutes be given, we may easily find all the other sides in the following manner.

Let the cosine of the arc of one minute, that is, the sine of the arc of 89 deg. 59', be called Q; and make the following analogies ; Å : 2 Q:: Sin. 2 : S. 1' + S. 3'. Wherefore the sine of 3 minutes will be given. Also, R. : 2 Q:: S. 3' : S. 2' +S. 4. Wherefore the S. 4' is given. And R.: 2 Q:: S. 4' : S. 3'+S. 5'; and so the sine of 5' will be had. Likewise, R.: 2Q:: S. 5': S. 4' +S. 6'; and so we shall have the sine of 6. And in like manner, the sines of every minute of the quadrant will be given. And because the radius, or the first term of the analogy, is unity, the operations will be with great ease and expedition calculated by multiplication, and contracted by addition. When the sines are found to 60 degrees, all the other sines may be had by addition only, by Cor. 1. Prop. 6.

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