In like manner (Art. 35.) the chord of the supplement of

3S. The sine, co-sine, tangent, secant, &c. of any are AB of a circle, whose radius is CA, is to the sine, co-sine, tangent, secant, &c. of a similar arc DE, whose radius is CD, as CA to

CD.

From the point B let fall BF perpendicular to CD (12. 1.), and through A, E, and D,

draw AK, EG, and DT, paral

E

K

B

lel to BF (31. 1.), then will BF be the sine of the arc BA, CF its co-sine, AK its tangent; EG the sine of ED, CG its co-sine, and DT its tangent (Art. 12. 16.); and since AB and DE each subtend the common angle at the centre C, they are similar, that is, they contain each the same number of degrees (part 8. Art. 239.); now since the angles at F, A, G, and D, are right angles, and the angle at Ccommon, the triangles BCF, KCA, ECG, and TCD, are similar

TAG D

(32. 1.), and have the sides about their equal angles proportionals (4.6.); that is,

First, FB: BC:: GE: EC, and alternately (16.5.) FB: GE:: BC: EC; that is, sine of arc BA: sine of arc ED :: rad. (BC) of the former arc : rad. (EC) of the latter.

Secondly, FC: CB:: GC: CE, and alternately FC: GC:: CB: CE; that is, co-sine of arc BA: cosine of arc ED :: rad. (CB) of the former : rad. (CE) of the latter.

Thirdly, KA: AC :: TD: DC, and alternately KA: TD :: AC: DC; that is, tang. arc BA: tang, arc ED:: rad. of BA: rad. of ED.

Fourthly, KC: CA: TC: CD. alternately KC: TC:: CA: CD; that is, secant of arc BA: secant arc ED :: rad. of BA: rad. of ED.

Fifthly, Because BC: CF:: EC: CG by conversion (prop. E. 5.) BC: FA:: EC: GD,

inversely (prop. B. 5.),

alternately FA : GD :: BA: versed sine of arc

FA: (BC=) AC :: GD: (EC=) DC AC: DC; that is, versed sine of arc ED:: rad. of BA : rad. of ED. Wherefore the sines, co-sines, tangents, secants, and versed sines of a given angle in different circles, are respectively as the radii of those circles. Q. E. D.

Hence, if sines, co-sines, tangents, &c. be computed to a given radius, they may be found to any other radius, by the above proportions.

39. The co-sine of any arc, is equal to half the chord of the Supplement of double that arc.

Let AE be an arc, C the centre, join CE, and from 4 draw AL perpendicular to CE (12.1.), and produce it to B, join BD, and from the centre C draw CF perpendicular to BD, ': DF=

lelogram, and CL=FB (34.1.); that is, the cosine of the are AE is equal to half the supplemental chord of (AB) double of AE. Q. E. D.

40. The chord of an arc is a mean proportional between its versed sine and the diameter.

Draw BK at right angles to DA (12. 1), then because DBA is a right angle (31.3.), DA : AB :: AB : AK (cor.8.6.); that is, the diameter is to the chord of the arc AEB, as the saine chord is to the versed sine of AEB. Q. E.D.

41. The sum of the tangent and secant of any arc, is equal to the co-tangent of half the complement of that arc.

Draw CH at right angles to DA (12. 1.), and let AE be any arc, AS its tangent, CS its secant, and the arc EH its complement. Bisect EH in B (30. 3.), and draw CBT meeting AS produced in T.

Then AT is the tangent of the arc AEB (Art. 16.) that is, the co-tangent of HB (Art. 17.) which is half the complement of AE.

Because AT and CH are parallel, the angle HCB=CTA (29.1.); but HCB=BCE ·· BCE=CTA ·: SC=ST (6.1.) AS+SC=AT; that is, the sum of the tangent and secant of the arc AE is equal to (AT) the co-tangent of (HB) half the complement of AE. Q. E. D.

42. The radius is to the co-sine of an arc, as twice the sine to the sine of double that are.

Because the right angled triangles ALC. 4KB have the angle at A common, they are equiangular (32. 1.), AC:

CL:: AB: BK, that is radius: co sine of AE:: twice the sine of AE sine of double of AE. Q. E. D.

THE INVESTIGATION OF FORMULÆ, NECESSARY FOR THE CONSTRUCTION OF THE TRIGONOMETRICAL CANON.

43. The sines and co-sines of two unequal arcs being given to determine the sine and co-sine of their sum and difference.

Let KF, FE be two unequal arcs of which the sines and co-sines are given, and let KF be the greater, from which cut off FD=FE the less (34.8.), join ED, and from the centre C draw CF perpendicular to ED (12. 1.) ·.· EL=LD (3.3.); draw DH, FG, LO, EM, each perpendicular to the diameter CK, and DS, LN each parallel to it (31.1.) meeting LO, EM in the points S and N.

Because EL=LD_EF=FD, ·: (30. 3.); and because LN is parallel to DS, the angle ELN=LDS (29. 1.), ... the right angled triangles ELN, LDS having all their angles equal, and the homologous sides EL, LD equal, are equal and similar (26. 1. and def. 1. 6.), ·.· EN=LS and NL-SD; also in the parallelograms NMOL, SOHD, we have NM=L0, NL=MO, DH SO, and SD=OH (34. 1.), ·.· NL=MO=SD=0H. Let the arc KF=A, the arc FEB, and the radius CF=R; then will the arc KE=(KF+FE=) A+ B, and the arc KD=(KF

Because NL is parallel to CO, and FG to LO and the angles at G, O, and N right angles, the triangles CFG, CLO, and ELN are equiangular (29 and 32. 1.), consequently (4.6.)

But EM (=MN+ EN=LO+EN), or sin A+B=

sin A. cos B+cos A. sin B

R

CM (=CO-MO=CO-LN), or cos A+B=

cos A. cos B-sin A. sin B

DH (=SO=LO-LS-LO-EN), or sin A-B= sin A. cos B-cos A. sin B

R

CH (=CO+OH=CO+LN), or cos A—B=

eos A. cos B+sin A. sin B

R

44. These formula for the sines and co-sines of the arcs A+B which are, it is plain, adapted to any radius R, may be simplified and rendered more convenient by putting R=1; they will then become

Formula 1. Sin A+B=sin A. cos B+cos A. sin B.

2. Cos A+B=cos A. cos B-sin A. sin B.

3. Sin A-B=sin A. cos B-cos A. sin B.

4. Cos A-B=cos A. eos B+sin A. sin B.

45. To find the sine and co-sine of multiple arcs, that is, if A be any arc, to find the sine and co-sine of nA.

Add the first and third of the above formulæ together, and in the sum let A be substituted for B, and B for A, and we shall have

sin B+A+sin B-A=2 cos A. sin B, that is, sin B+A=2 cos A. sin B-sin B—A. (Y).