PROPOSITION XI. (T)The right angled triangles FGC, TBC, CAK, CEI, are equiangular and similar. For FGC, TBC are right angles, and the angle FCB=TCB, therefore the remaining angle CFG=/CTB. The triangle CEF triangle CGF, for the CFG = /FCE (Euclid I and 29), and FEC= CGF, being each of them right angles, and the side CF is common to both the triangles, therefore they are equal (Euclid I and 26). 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. (Z) CB : BT :: AK : CA. radius tangent co-tangent radius. (A) {co-tangent: EF CG :: CK : CF. cosine co-secant : radius. (B) By comparison, Cosine sine: co-tangent: radius. (U and Z, above.) tang. :: co-secant : radius. (W and Y, above.) Secant rad. :: co-secant : co-tang. (X and A, above.) Secant = (C) Tangent X co-tangent radius square. (Z, above.) Therefore the tangent of any arc x its co-tangent the tangent of any other arc × its co-tangent. (D) Sinex co-secant radius square. (Y, above.) Therefore the sine of any arcx its co-secant sine of any other arc its co-secant. (E) Co-sine secant radius square. (X, above.) Therefore the co-sine of any arc x its secant co-sine of any other arc x its secant. (F) Square radius=square sine + square co-sine. For, CF2=GF2+CG2 Euclid I and 47.) (G) Square (G) Square radius=square secant-square tangent=square cosecant-square co-tangent. 2 For, CB2 CT-TB2 (Euclid I and 47), and ▲c2=CK2- AK2. (H) Square radius=co-sine x secant (X. 97.)=sine × co-secant (Y. 97.)=tangent × co-tangent (Z. 97.) tangent X co-sine (I) Sine = radius -(U.97) — tang. × rad(W.97)= secant square rad cosine x rad (Y.97.)= (B. 97), &c. Co-sec co-tangent (K) And, generally, if A=any arc*, rad radius, cos=cosine, tang tangent, cot co-tangent, sec=secant, cosecco-secant, vers=versed sine, vers sup-versed sine of the supplement, or the superversed sine, the following formula will be easily deduced, where the sign × is represented by a point (.). (L) I. Sine A=√ rad2 — cos2 A = -- COS A.tang A tang A.rad sec rad2 CoSec A vers sup A.vers A (2 rad.vers A)-vers2 A. (M) II. Cos Arad rad/(2rad.vers A)-vers A_rad/(2rad.vers sup A)-vers sup2 A * Emerson's Trigonometry, 2d edit. Prop. I. Scholium. =COS (P) V. Sec Arad + tang2 A= rad.cosec A cot A rad. cos rad2 rad3. sine A cos A. tang2 A (vers sup A-rad).rad √(2 rad.vers A)-vers2 A √(2 rad.vers sup A)-vers sup2 A rad3 sine A.Cot rad2 rad2sine2 A cot A rad2 rad tang A tang A. cot A COS A rad2 + cot A= (R) VII. Vers A=2 rad-vers sup Arad-cos Arad rad2 sine2 A rad rad.cot A √sec2 A rad2 /rad2+cot2 A rad2 rad = rad rad rad + ==rad+ rad sine A+ tang A =rad+ 'cosec2 A-rad2= Cosec A tang A sine A.cot A rad2 cot A+ cosec A rad rad + =rad+ rad. rad (T) If the co-versed sine be wanted, it may be found by subtracting the sine from radius; that is, co-vers A = rad sine A; also the chord vers2 A sine2 A= √rad—cos A)2 +(rad2 — cos2▲)=√✓/2 rad.(rad—cos a). (U) Besides the preceding formulæ, others may be deduced, thus, sine A+ cos Arad2 (F. 97) rad2+tang2 A sec2 A, and rad2 + cot2 A = cose2 A (G.98.). Now because tang A = rad. sine A rad2.sine2 A hence rad2+ tang2 ▲ Cos2 A (N.98.) tang2 A= rad2.sine2 A rad2. (cos2 A+ sine2 A) rad4 COS A rad2 + &c. GENERAL PROPERTIES OF SINES, TANGENTS, &c. OF DOUBLE ARCS AND OF HALF ARCS. PROPOSITION XII. (Plate I. Fig. 1.) (W) The right angled triangles bGF, BGF, bFB, and CZB, are equiangular and similar; and cz, the co-sine of the arc Bi, is equal to the half of be, the chord of the supplement of double the arc Bi. For bGF, BGF, and bFB, have already been shewn to be equiangular (P. 96), and the triangles CZB, FGB, have the angle at B common to both of them; also the angle bFB, being an angle in a semicircle, is a right angle; and since cz is parallel to bF, by construction (R. 96), the angle czB is likewise a right angle. Now BZ: CZ:: BF: Fb, but Bz = BF (R. 96), therefore Cz=16F. 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; versed sine supplement of double the arc. CB: CZ:: BF (=2bz): GF radius: cosine of an arc :: double the sine: sine of double the arc. (X) (Y) (Z) (A) (B) (C) double the sine of an arc sine double the arc :: double the cosine : versed sine of double the arc. BZ: CZ:: GB GF (D) sine of an arc: its cosine :: versed sine double arc: sine double arc. (E) BZ. 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): 2GF :: bf. (=2cz): bG (E) (F) sine of an arc its cosine :: sine double arc: versed sine supplement of double arc. BG GF GF: bG versed sine double arc : sine double arc : : sine double arc versed sine of the supplement of the double arc. (G) By comparing these proportions with the proportions Prop. XI. and casting out the equal terms, various other proportions may be formed. Thus (W. 97.) it is said radius sine secant : tangent, and (X. 100.) we have : radius: sine :: double sine: versed sine double arc, therefore secant : tangent :: double the sine: versed sine double the arc, &c. for others. (H) COROL I. The rectangle of the radius and the sine of any arc, is equal to double the rectangle of the sine and cosine of half that arc. For (A. 100) CE: CZ:: 2BZ: GF Therefore CB X GF2BZ X CZ. (I) COROL II. The rectangle of radius and half the versed sine of an arc, is equal to the square of the sine of half that arc. For (Y. 100) CB: BZ:: BF: BG And CB: BZ: BF: BG That is CB: BZ :: BZ: 1BG Therefore CB × BG=EZ2; now CB is the radius, BG the versed sine of BF, and вz is the sine of Bi, which is half the arc BF. (K) COROL III. The rectangle of radius and half the versed sine of the supplement of an arc, is equal to the square of the cosine of half that arc. Therefore CB bocz2. Where CB is the radius, bo the versed sine of the supplement of the arc BF, and cz the cosine of half the arc BF cosine of the arc Bi. (L) COROL IV. The diameter of the circle is to the versed sine of any arc, as the square of the radius is to the square of the sine of half that arc. For, (I. 101.) CBX BGBZ, multiply by the radius CB, therefore CBBG:: CB2: BZ2 that is 2CB: BG:: CB2: BZ2 (M) COROL V. The rectangle of the sine of any arc, and of the cotangent of its half, is equal to double the square of the cosine of half that |