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so the Sine, Tangent, or Secant of any Arc, may be bad by Help of this Table; and contrarywile, a Sine, Tangent, or Secant, being given, we may find the Arc it expresses. Take Notice, That in in the following Tract, R signifies the Radius, S a Sine, Cof. a Cofine, Ta Tangent, and Cot, a Cotangent.
The CONSTRUCTION of the Tri
The two Sides of any Right-angled Triangle being
given, the oiber Side is also given. OR (by 47. of the first Element) AC=AB9
+ BCq and B and interchangeably Acq-AB=BCq. Whence, by the Extraction of the square Root, there is given ACEVABq+BCand AB=VACY-BCq. And BC=VACY-ABq.
cp, being given, to find the Cofine DF.
in the Right-angled Triangle CDE, there will be given (by the last Prop.) VCDq- DEq=DF.
PROBLEM. The Sine DE of any Arc DB being given, to find
DM or B M tbe Sine of half the Arc. DE being given, CE (by the last Prop.) will be given, and accordingly E B which is the Difference between the Cofine 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=į the Arc B D.
PROBLEM. The Sine B M of the Arc BL being given, to find
the Sine of double that Arc.
HE “Sine B M being given, there will be given
(by Prop. 2.) the Cofine CM. But the Triangles CBM, DBE, are equiangular, because the Angles at E and Mare Right Angles, and the Angle at B common. Wherefore (by 4. 6.) we have OB : CM :: BD, or 2 BM:DE. Whenee, fince the three first Terms of this Analogy are given, the fourth also, which is the Sine of the Arc DB, will be known.
Coroll. Hence, CB : 2 CM::BD: 2 DE, that is,
the Radius is to double the Cofine of one half of the Arc D B, 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:: 1 CB: CM. Wherefore the Sine of any Arc, and the Sine of its Double being given, the Cofine of thie Arc it self is given.
PROPOSITION V. The Sines of two Arcs BD, FD, being given, to
find F I the Sine of the Sum, as likewise EL, the Sine of their Difference. I ET the Radius CD be drawn, and then CO is
the Cosine of the Arc FD, which accordingly is given, and draw OP thro? 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 confequently is known, Also we have CD:CK::FO:FM, and fo likewife this shall be known. But because FO= EO, then will FMMG=ON. And so OP +FM FI=Sine of the Sum of the Arcs: And OP-FM, that is, OP-ONEEL=Sine of the Difference of the Arcs. W. W. D.
Coroll. Because the Differences of the Arcs BE, BD,
BF, are equal, the Arc B D shall be an Arithmetical Mean between the Arcs BE, BF.
PROPOSITION VI. The fame Things being supposed, Radius is to dou
ble the Cofine of the mean Arc as the Sine of the Difference, to the Difference of the Sines of the Extremes.
OR we have CD:CK::FO:FM, whence.
by doubling the Consequents CD: 2 CK :: FO: 2 FM, or to FG; which is the Difference of the Sines EL, FI. W.W.D.
Coroll. If the Arc BD be 60 Degrees, the Difference
of the Sines FI, EL, shall be equal to the Sine, FO; of the Distance, For in this Case, CK is the Sine of 30 Degrees, Double thereof being equal to Radius; and fo fince CD= 2 CK we shall have FO=FG. And consequently, if the two Arcs BE, BF, are Equidistant from the Arc of 60
Degrees, the Difference of the Sines shall be equal to the Sine of the Distance FD.
Hence, if the Sines of all Arcs be given distant from one another by a given Interval, 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 > Sine of 59 Degrees +Sine of i Degree. And the Sine of 62 Degrees =Sine of 58 Degrees + Sinę of 2 Degrees. Also the Sine of 63 Degrees = Sine of 57 Degrees + Sine of 3 Degrees, and so on.
If the Sines of all Arcs, from the Beginning of a Quadrant to any Part of the Quadrant, distant from each other, by a given Interval be given, thence we may find the Sines of all Arcs to the Double of that Part. For Example, Let all the Sines to 15 Degrees be given; then, by the precedent Analogy, all the Sines to 30 Degrees, may be found. For Radius is to double the Cofine of 15 Degrees, as the Sine of 1 Degree, is to the Difference of the Sines of 14 Degrees, and 16 Degrees; fo alfo is the Sine of 3 Degrees, to the Difference between the Sines of 12 and 18 Degrees; and so on continually until you come to the Sine of
30 Degrees. After the fame Manner, as Radius is to double the : Cofine of 30 Degrees, or to double the Sine of 60
Degrees, so is the Sine of 1 Degree to the Diffe-
Fig. for the De FINITION S.
Rad..Fan. Co-fine Sine
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 Be
ginning of the Quadrant be found, by having the Šines and Cofines of one and two Minutes given. For as the Radius is to double the Cofine of 2 : ; Sine i': Difference of the Sines of i' and 3':: Sine z': Differenee of the Sines of o' and 4', that is, to the Sine of 4'. And so the Sines of the four firft Minutes being given, we may thereby find the Sines of the others to 8', and from thence to 16, and so on.
THE OR E M.
In small Arcs, the Sines and Tangents of the same
Arcs are nearly to one another, in a Ratio of
OR because the Triangles CED, CBG, are
equiangular, CE:CB :: ED :: BG. But as the Point E 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
I equal to BG. If EB be less than the
10,000,000 Part of the Radius, then the Difference between the Sine and the Tangent will be also less than the
10,000,000 Part of the Tangent.
Coroll. Since any Are is less than the Tangent, and
greater than its Sine, and the Sinę and Tangent of a very small Arc, are nearly equal; it follows that the Arc shall be nearly equal to its Sine; and so in very small Arcs it shall be, as Arc is to Ars, fo is Sine to Sine.