1

And as A, in trigonometry, must always be less than 180°, A will always be less than 90°, and consequently its sine, tangent, &c. will have the same sign as in the table in art (200).

GENERAL PROPERTIES OF THE SINES, TANGENTS, CHORDS, ETC. OF SINGLE ARCS.

PROPOSITION VIII. (Plate I. Fig. 1.)

(205) The chord of any arc is a mean proportional between the diameter and the versed sine of that arc.

The angle bгB being in a semicircle, is a right angle (31 Euclid III). And FG, by the definition of a sine, is perpendicular to bв, therefore the triangles bGF, and BGF, are similar to each other, and to the whole triangle bгв (8 Euclid VI). Hence,

bb: BF BF BG, where bв is the diameter, and BF the chord, and BG is the versed sine of the arc bif.

(206) COROL. The sine of any arc is a mean proportional between the versed sine of that arc, and the versed sine of its supplement.

For the FbGL BFG, consequently bG: GF :: GF : BG.

PROPOSITION IX. (Plate I. Fig. 1.)

(207) The square of the sine of any arc added to the square of its versed sine, is equal to the square of the chord of that arc; or to four times the square of the sine of half that arc.

For GF2+GB2BF2 (47 Euclid I).

Draw ci parallel to br, and it will cut BF at right angles in z; but if it cut it at right angles, it will bisect it (3 Euclid III), therefore BZZF, and BF2BZ, and consequently BF2=4BZ2.

PROPOSITION X. (Plate I. Fig. 1.)

(208) The tangent of any arc is a mean proportional between the sum and difference of the radius and the secant; viz. sec+rad: tan:: tan: sec-rad

Join FC, and produce it to h and to T, so as to meet the tan

gent BT of the arc BF.

2

Then hTX FT=TB2 (36 Euclid III), that is hT: TB TB: FT. (17 Euclid VI) But hт=sec CT+ rad, and FT=sec CT-rad; therefore sec+rad: tan :: tan : sec-rad.

PROPOSITION XI. (Plate I. Fig. 1.)

(209) The right angled triangles FGC, TBC, CAK, CEF, are equiangular and similar.

For FGC, TBC are right angles, and ▲ FCB= TCB, therefore the remaining CFG = CTB. The triangle CEF triangle CGF, for the CFG = FCE (29 Euclid I), and FEC= CGF, being each of them right angles, and the side cF is common to both the triangles, therefore they are equal (26 Euclid I). Again, AK is parallel to EF, by the definition of a tangent and sine: therefore the triangle CAK is equiangular with the triangle CEF; and consequently with CGF and CBT.

sec : rad :: cosec : cot (C) and (F).

(211) From (E) tan. cot=rad2.

Therefore the tangent of any arc x its cotangent=the tangent of any other arc x its cotangent.

(212) From (D) sin. cosec=rad2.

Therefore the sine of any arc x its cosecant sine of any other arc x its cosecant.

(213) From (C) cos. sec=rad2.

Therefore the cosine of any arc x its secant cosine of any other arc x its secant.

(214) Also rad2 sin2 + cos2.

For, CF2GF2+cG2 (47 Euclid I).

=

(215) rad2sec2-tan2=cosec2-cot

For, CB2CT2-TB2 (47 Euclid I), and Ac2=CK2-AK2. (216) rad2=cos. sec (by C art 209)=sin. cosec (by D) tan. cot (by E).

(217) sin =

tan. cos
rad

(by A) =

tan. rad

rad

(by B)='

(by D)

(218) Let A denote any arc*; then adopting the notation

in arts (107) and (203), we have

I. sin Arad2 — cos2 A=

COS A tan A
rad

rad. tan A

✓rad2 + tan2 A

cot A

rad

tan A

sec

rad 2

CoSec A

✔suvers A. vers A= √√2 rad. vers a—vers2a.

rad/2rad. vers A-vers2 A rad✔(2rad. suvers a) -suvers2 A

* Emerson's Trigonometry, 2d Edit. Prop. I. Scholium.

VIII. suvers A=2 rad-vers Arad + √rad2-sin2 A =

(219) The coversed sine may be found by subtracting the sine from radius; that is, covers A-rad-sin A; also the chord = V vers2A + sin2 A = √(rad-cos A)2 + (rad2 — cos2 A)= √2 rad (rad-cos A).

(220) Besides the preceding formulæ, others may be de

duced, thus, sin2 A+ cos 2 Arad 2 (214), rad2+tan2 A= sec2 A, and rad2+ cot2 A=cosec2 A (215). 2 Now because tan A =

rad2. sin2 A

hence rad2+ tan2 A

COS A

COS2 A
2

GENERAL PROPERTIES OF SINES, TANGENTS, ETC. OF DOUBLE ARCS AND OF HALF ARCS.

PROPOSITION XII. (Plate I. Fig. 1.)

(221) The right-angled triangles bGF, BGF, bFB, and CZB, are equiangular and similar; and cz, the cosine of the arc Bi, is equal to the half of bF, the chord of the supplement of double the arc Bi.

For bGF, BGF, and bFB have already been shown to be equiangular (205), and the triangles CZB, FGB, have the angle at B common to both of them; also the angle bгB, being an angle in a semicircle, is a right angle; and since cz is parallel to br, by construction (207), the angle czB is likewise a right angle. Now BZ: CZ:: BF: Fb, but BZ = BF (207), therefore cz=bF.

Hence the following proportions.

CB: BZ (sine arc Bi) :: BF(=2Bz) : BG

radius: sine of an arc :: double that sine: versed sine of double that arc.

CB: Bz :: bf (=2cz): GF

radius: sine of an arc :: double its cosine: sine double

the arc.

CB: cz :: bf (=2cz): bG

radius: cosine of an arc:: double its cosine : su

versed sine of double the arc.

CB; cz :: BF (=2bz) : gf

(A)

(B)

(C)

(D)

radius: cosine of an arc :: double the sine : sine of double the arc.

(E)

(F)

(G)

BF (2BZ) BG:: bf (=2cz): GF

double the sine of an arc: versed sine double arc :: double cosine : sine of double the arc.

BF (=2BZ) GF :: bf (=2cz): bo

double the sine of an arc: sine double the arc :: double the cosine: suversed sine of double the

arc.

BZ CZ: GB GF

sine of an arc its cosine: : versed sine double arc : sine double arc.