« السابقةمتابعة »
than 10, but the logarithmic tangents, secants, angle A F D must also be half a right angle; and corangents, and cosecants admit of all possible therefore A D and A F are equal, or the tangent values.
of 45° is equal to the radius. On Tue Signs of TrigonOMETRICAL Quan Prop. II.--The secant of an arc is equal to
the sine of its tangent and the tangent of half its When geometrical quantities that are measured complement. from a given point or line are considered analyti Let A B, fig. 5, be any arc, AD its cally, they are considered as positive or negative, tangent, and C D its secant; produce A D till and are accordingly designated as + or - ac DE is equal to DC; join C E, and draw DHI cording as they lie on the same side or on op- a perpendicular on CE. Then as the right angled posite sides of the same point or line. Thus in triangles CAE and DHE have the common fig 2, the sines are estimated from the di- angie E and the right angles CAE and DHE ameter A A', and in the semicircle A E A' they equal; the remaining angle AC E in the one is are considered as t; but, as in the other semi- equal to the remaining angle E DH in the other. circle A E A they fall on the other side of the But the triangles EDC being isosceles, the angle diameter AA', they are then considered as — D is bisected by the perpendicular DH; thereThe cosines, being estimated from the centre C, fore the angle ACE is equal to half ADC, or are considered as + in the first quadrant A E; to half the complement, of AC D. Hence A D but as in the second quadrant EA', and the third the tangent of AC D, and A E the tangent of A E', they fall on the other side of the centre Chalf its complement, are equal to C D its sethey are then considered as —; and again in the fourth quadrant E'A they become +, as in the Prop. III.-If A represent the greater and B
the less of two arcs, it is proposed to investigate first quadrant. As tan. = tan. is + in the the relations between the sines and cosines of first and third, and in the second and fourth
those arcs and the sines and cosines of their
1 quadrants; and, as cot. = the tan. and co Let K, fig. 6, be the centre of the circle,
CE the greater arc A, and E F or ED the tan. have the same sine; and as sect. = less arc B; join F, D, and K, E, which will bi
sect. F D in Il and cut it at right angles. Draw and cosect. =
the sect. and cos. have the FL, HM, E N, and D O perpendicular to KC same sine, and the cosecant and sine have the and GH, ID parallel to KC. Then the trisame sign.
angles F G H and HID are obviously identical, Prop. I.— The chord of 60° and the tangent GF and H I being equal; and GH, ID, LM, of 45° are each equal to the radius; the side of and M O are also equal. If from the right angles 30°, the versed sine of 60°, and the cosine of 60°; taken, the angles FHG and K H M will remain
FH K and GHM the common part GHK be are each equal to half the radius; and the secant of 60° is double the radius.
equal; and as the angles at G and M are also Let D, fig. 4, be the centre of a circle, equal, being right angles, the angles HK M AB an arc of 60°, and AC an arc of 45°; and HFG are equal, and consequently the trijoin A B, B D, and from B draw B E perpendi- angle F G H, which is identical with the triangle cular to AD, and from D draw DG H perpen- HID, is similar to K HM, which again is evidicular to A B, and it will bisect both the chord dently similar to K E N. A B and the arc ACB. Draw the tangent AG
Now E N is the sine, and K N the cosine of the meeting the secants DBG and DCF in G and greater arc A, FH the sine and K H the cosine F. Now, as the angle B D A is 60°, the sum of of the less arc B; FL the sine and KL the cosine the two equal angles D A B and D B A is 120°; of the sum of the arcs A + B; DO the sine and therefore each of these angles is 60°, and conse K O the cosine of the difference of the arcs A-B. quently the triangle A B D is equilateral: whence And LF = }M + CF; DO = HM - HI A B, the chord of 60° is equal to the radius. = H M - G F; KL = KM – G H, and KO And as, in the equilateral triangle A BD, the KM +ID=KM + GH. perpendicular B E bisects A D, and the perpen From the properties of similar triangles we dicular DG H bisects both the chord A B and have K E:EN::KH: HM; or rad. : sin. A the arc AHCB, D E the cosine of 60°, E A the
:: cos. B:H M; whence if radius be unity HM versed sine of 60°, and B C the sine of 30°, are = sin. A. cos. B. And K E:KN:: FH: each equal to half the radius. And by similar FG; or rad. : cos. A :: sin. B : FG; whence triangles DE: DB::DA:DG; whence, as F G = cos. A sin. B. And K E:KN::KH DB = 2 D E, DG= 2 D A, or the secant of :KM; or rad. : cos. A :: cos. B : KM; whence 60° is double the radius.
KM = cos. A cos. B. Lastly, K E:EN:: Again as the angle DAF is a right angle, FH :HG; or rad. : sin. A:: sin. B: HG, and the angle ADF is half a right angle, the whence HG sin. A sin. B. (I.) Hence FL, sin. A + B = = sin. A cos. B + cos. A · sin. B.
DO, sin. A B = sin. Acos. B cos. A·sin. B.
B = cos. A cos. B + sin. A · sin. B. By taking the sum and the difference of the first tw of these four equations, and also of the last two, we have
(II.) sin. A + B + sin. A B = 2 sin. A: cos. B.
sin. A + B - sin. A B = 2 cos. A sin. B.
cos. A - B - cos. A + B = 2 sin. A · sin. B. On the four equations which we have now obtained may be remarked that A is half the sum of A + B and A
B, and that B is half the difference of A + B and A – B; if therefore instead of A + B we were to put A', and instead of A.- B we were to put B', we must put for A A' + B'
A - B in the same equations
and for B
,and the equations would then stand in this form: 2
A' + B' A' - B (III.) sin. A' + sin. B' = 2 sin.
A' + B' A B' sin. A' sin. B' = 2 cos.
A' + B' A' B' cos. A' + cos. B' = 2 cos.
If in these last expressions the arc B’ were to be considered as evanescent, we should have
whence sin A+B= 2 sin.
1 + cos. A 1 + cos. A' = 2 cos.? ; whence cos. 2
2 A' 1- cos. A' 2 sin.? ; whence sin.
- cos. A 2
A + B A + B A - B (V.) Again, sin. A + sin. B · sin. A -- sin. B= 2 sin.
2 sin. 2
= sin. A + B sin. A B. Whence sin A 2
B:sin. A- sin. B :: sin. A + sin. B:
: sin. A + B.
2 Proceeding in the same way we have A+B А. -B
A+B 2 sin.
tan. sin. A + sin. B
rad. 2. cos.
(VII.) If B, in each of these expressions, vanish, we have
sin. A tan, } A From the first equation the identical expression
tan. Í A A
tan. sin. A
2 From the fourth the same expression as from the second
1 + cos. A
(VIII.) Transposing the first and second equations (II.) we have
sin, A + B = 2 sin. A cos. B - sin. A - B
sin. A + B 2 cos. A · sin. B + sin. A - B’ and recollecting that it has been shown in proposition I. that sin. 30° and cos. 60° are each half radius; if we consider A in the first of these equations as 30°, and in the second as 60°, the formula will become
IX. sin. 30° + B = cos. B — sin. 30° B
sin. 60° + B = sin. B + sin. 60° В'
B 1 + sin. B : = 2 sin. 45° +
= 2 sin.? 45° + = 2 cos.? 45°
B 1-sin. B= 2 cos. 45° + sin. 45°
= 2 cos.' 45° + = 2 sin.: 450 2
tan.? 45° + cot.45°
A 2 cos.?
Prop. IV.-A representing the greater, and B the less of two arcs, it is proposed to investigate the relation between the tangents of those arcs and the tangents of their sum and their difference.
sin. A + B tan. A + B (I.) By formula 1, proposition III., we have
cos. A + B
rad. sin. A.cos. B cos. A. sin. B
+ sin. A. cos. B + cos. A . sin. B
cos. A.cos. B tan. A + tan. B cos, A. cos. B - sin. A. sin. B cos. A.cos. B sin. A. sin. B
tan. A, tan. B cos. A.cos. B cos. A.cos. B sin. A В B. tan. A
tan. B and similarly we have
Hence, radius being cos. A B
1 + tan. A. tan B
tan. A + tan. B unity, we have tan. A + B =
1 F tan. A . tan. B (II.) If in this expression A = 45°, then tan. A = rad. (proposition I.), and we have tan.
1 + tan. B 45° + B =
1 + tan. B
If t, t', t", &c., be the tangents of the arcs A, B, C, &c., then, considering A + B as one arc,
tan. A + B + tan. C we have tan. A + B +C=
But tan. A + B =
t + ť tan. A + B . tan. C
1-tit t + ť
+ to - t t
t +ť +ť" - tť " (III.) Hence tan. A + B +C = t +ť
And if A + B 1 - (tľ + + + + + t') 1
+C = 180°, which is the case when A, B, and C are the angles of a plane triangle; then, since tan. 180o = 0, the theorem gives t + R + ' = tť t";(IV) or when radius is unity, the sum of the tangents of three arcs, which together make 180°, is equal to the continued product of those tangents.
Prop. V.-From the sine and cosine of any given arc A, it is proposed to investigate expressions for the sines and cosines of its multiples.
In formula 1, proposition III., we have sin. A + B = sin. A.cos. B + cos. A . sin. B.
sin. 3 A = sin. A, cos. 2 A + cos. A. sin. 2 A.
sin. 4 A = sin. A . cos. 3 A + cos. A . sin. 3 A, &c. Again, in the same proposition, we have cos. A + B = cos. A.cos. B — sin. A . sin. B, and substituting for B, A, 2 A, 3 A, &c., in succession, we have (II.) cos. 2 A = cos.? A — sin. A = 2 cos.: A – 1=1 — 2 sin.? A.
cos. 3 A = cos. A.cos. 2 A-sin. A . sin. 2 A.
cos. 4. A = cos. A.cos, 3 A sin. A . sin 3 A, &c. Again, by formula 8, proposition III,
sin. A + B = 2 sin. A . cos. B — sin. A - B. If in this formula B=n 1. A, it becomes (III.) sin. n A = 2 cos. A . sin. n 1 . A- sin. n
- 2. A. By adding together the third and fourth expressions, formula 2, proposition III., and transposing, we have cos. A + B = 2 cos. A . cos. B. cos. A – B.
Here let B=n - 1. A, and the formula becomes (IV.) cos. n A = 2 cos. A . cos. n — -1A
2. A. : If in formula 1 and 2 of this proposition we substitute 1 - cos. ? for sin., and W1 – sin. for cos., the expressions will stand thus :
(V.) sin. A =s.
sin. 2 A = 2 S N 1 sa.
sin. 4 A = 48 8 so.n1 5, &c.
(VI.) cos. A=c.
cos. 2 A= 2.c 1.
cos. 4 A = 8C 8 C2 + 1, &c. A class of very elegant and curious formulæ inay be deduced from the last class by means of a particular substitution.
(VII.) Make 2 cos. A=;+ ; then 2 cos. 2 A=2.(2 cos. ?A — 1) = 2
214.5+= 1) 3*+ **
and a like substitution for 2 cos. 3 A, 2 cos. 4 A, &c., will give 2 cos. 3 A = 88+
2 cos. 4 A = 2 +
in . m -1 . m
sin, m A, and (cos. A N- 1. sin. A)" = formula required for the solution of the different
cases of right angled plane triangles :—Let cos. m A N - 1. sin. m A.
ABC (fig. 7,) be a plane triangle, right-angled Expanding these expressions, and adding them,
at B; round A as a centre with any radius,
AD, describe the arc DG; and from G and D we have cos. m A = cos. -A
draw G F and D E perpendicular to A B.
Then G F will be the sine of the angle A, 1.m
-2.m A. sin. 'At
A F its cosine, D E its tangent, and A E its se
2.3.4 4 A. sin. -A- &c.
cant; and, as the angle C is the complement of
A, the sine, tangent, and secant, of one of these If we subtract them, we have sin. m A
angles will be the cosine, coangent, and cosecant ---A. sin. A
cos. respectively, of the other. Now, by the properties 2.3
of right angled triangles AC-V A B? + BC”, -3 A . sin. 'A + &c. Prop. VI.-It is proposed to express the
A B = V(A C + BC).(AC-BC), and BC powers of the sine and cosine of an arc in terms =N (AC + A'B).(AC - AB). of the cosines and sines of the multiple arc.
Again, by similar triangles A C:CB::AG: Expressions of the kind which we have here GF, that is : : rad. : sin. A ; hence A C. sin. A
A C. sin. A proposed to investigate are of the greatest im- = BC rad; therefore BC= portance in all mathematical investigations con
rad. nected with physical astronomy.
A C cos. C.
B Crad. BC cosec A -, and ACE rad.
rad. As 2 cos. A = x + 2* , cos. "A
AC cos. A And similarly AB rad.
rad. zh AC. sin. C and AC=
A B•rad_AB . sec. A rad.
rad. tnx *-* + n
AB · cosect C
:: :D E, that : : rad:
tan. A ; therefore AB tan. A=BC rad.; hence +
BC rad. BC.cot. A BC. tan. C
AB= the terms equally distant from the middle term
tan. A rad.
A Btan. A A B cot.C of the series. But 2 cos. n A = 2." + BC=
Prop. VIII.-It is proposed to investigate the Hence 2" . cos. "A= 2. cos. n A tn. cos.
formulæ required in the solution of the different 1
cases of oblique-angled plane triangles. – 2 . A + n .
4 A +, &c. 2
Formulu 1.-AB. sin. A=CB .sin. C (figs. cos. " A = cos. n A tn. cos. n
8 and 9). For let BD, in each figure, be a 2
perpendicular from B on the base A C, or 4.A &c.
A C produced. Then, by the last proposition,
in the right-angled triangle CBD CB. sin C As examples of the formula
B D rad., and AB:sin. BAD-BD. rad.; If n = 2, 2.cos. A cos. 2 A +1
therefore AB .sin. BAD=CB: sin. C; and, n = 3, 2'. cos. 3A = cos.3 A + 3 cos. A sin. BAD (fig. 8) =
sin. BAC; whence gen=4, 2cos. ‘A = cos. 4 A + 4ros. 2 A + 3
nerally A B. sin. A= BC sin. C. Hence CB: n = 5, 24.cos.'A= cos. 5 A + 5 cos. 3 A + 10 BA:: sin. A:sin. C. cos. A, &c. To obtain a general form for sin. "A; in tan. A BD : tan. DBC, or :: cot. A: cot. C. For
Formula 2.- In the same figures AD:DC:: the expression for cos. "A, substitute - A, (9 (prop. VII.) B D. tan. A B D= AD rad., and signifying a right angle); then 2 * cos. * q-A
BD. tan. DBC=CD, rad. and therefore A D:
rad. :CD rad. ::BD:tan, ABD: BD , tan. = cos. nq-n A tn.cos. n - 2 (9-A) + n
DBC'; hence AD: CD::tan ABD: tan. cos. n -- 4.9-A, and by reduction we
DBC; or :: cot. A : cot. C.
Formula 3.—In any triangle a A, B C (fig. have + 2 *-1 sin."A-: sin. n A -n. sin. n 2.
10), in which A B and C represent the
angles, and a, b, and c, the sides opposite those A + n. sin. n --4.A , &c.
A + B A--B. angles, a + 6:0–
b:a-b::tan. -: tan. As examples of this form
2 If n = 2; 2 sin. 'A = cos. 2 A +1
For formula 1 of this proposition a : b :: sin. A: n=3; 22 sin. 3A = sin. 3 A + 3 sin. A sin. B; therefore a +bia-b:: sin. A + sin. n=4;23 sin. *A= cos. 4 A ---- 4 cos. 2 A +3 B : sin. A — sin. B :: (formula 1, prop. III.) tan.
n=5;2'sin. 'A=sin. 5 A -- 5 sin. 3 A + 10 A + B A-B sin. A,&c.
2 The three preceding propositions contain a Formula 4.-In any triangle, as A B C (fig. condensed view of what, in modern mathematics, 9), if B D be a perpendicular, let fall from has been termed the arithmetic of sines.
the vertex upon the longest side, then AC: PROP. VII.-- It is proposed to investigate the AB + BC: AB-BC: AD-DC. For