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(17.) Trigonometrical, like all other algebraical at Ca' the cosine vanishes, and the sine coincides quantities, are susceptible of different signs under with Ca', and ... =-1. different circumstances. The signs of the sine and Through a'CA the sine diminishes continuing cosine are determined by the same rules as the co negative, and the cosine increases and is positive, ordinates of a point in analytic geometry.

thus changing its sign in passing through zero at plain the several changes of sign of the sine and co Ca'. sine, let C be the centre of a circle whose radius Thus the several changes which the sine and coCA is unity, and let CA be the initial position of sine undergo in one revolution of the radius are the revolving radius.

evident, and they suffer the same changes every When the radius coincides with CA, the sine of revolution. the arc = 0, and the cosine = 1. This will be mani. By the formulæ found in (10) and (16), it follows fost from considering the triangle CPM to be con that the tangent and cotangent are positive when tinually changed by the radius CP approaching the sine and cosine have like signs, and negative CA. During this change PM, the sine of the an when they have unlike signs; and by (12) it apgle continually diminishes, and CM its cosine con- pears that the sign of the secant is always that of tinually approaches to equality with CA or unity; the cosine; and by (13) that the sine of the coseand when the angle at C actually vanishes, and CP cant is always that of the sine. Thus the signs of coincides with CA, then PM vanishes, and CM the sine and cosine regulate those of all other trigbecomes equal to CA.

onometrical terms.
Hence for all angles terminated by the radius
CA, the sine 0, and the cosine : 1.
While the revolving radius moves from CA

SECTION II. through the angle A Ca, the sine PM continually increases, and the cosine decreases until it coin- of the Relations between Angles and their Sums and cides with Ca, where the sine coincides with Ca,

Differences. and is ::: 1, and the cosine vanishes, or = = 0. Through the angle ACa we shall consider the find the sines and cosines of their sum and difference.

(18.) Given the sines and cosines of two arcs to sine and cosine, both positive, and, in general, we shall consider those sines which are measured from

C

Let AOB be the the diameter AA' in the direction Ca as positive, and those which are measured in the opposite di

greater angle, a and P

BOC the less'. Then rection Ca' as negative.

AOC will be the sum Also, those cosines which are measured in the

M

wtor or the differ. direction CA we shall consider positive, and those

ence 2 according in the opposite direction CA' negative. It will be

as OC and OA lie at found that by this arrangement all trigonometrical

0

different sides of OB quantities will change their signs upon passing

NE

or at the same side. through zero and infinity. As the radius revolves

B Take any point P on through the angle aCA' from Ca towards BA', the

OC and draw PM sine diminishes, but is still positive, and the cosine

D

and PN perpendicuincreases, and is negative. Thus, at Ca the cosine

lar to OB and OA, passes through zero, and changes its sign. When

and from M draw the radius coincides with CA', the sine 0, and the

ME perpendicular to cosine coinciding with CA' is = -1.

0


Ε Ν

OA. Through M Through the angle A'Ca' the sine increases, and

draw MD parallel to OA to meet NP produced if the cosine diminishes, both being negative, the sine

necessary at D. By the definition we have changing its sign in passing through o at CA'; and

ME
PM

PN at Ca' the cosine vanishes, and the sine coincides

sin.a's sin.(-+')= MO'

PO
with Ca', and ::: =1.
Also, those cosines which are measured in the

EO
MO

NO

cos.(-+')= direction CA we shall consider positive, and those

Мо?
PO

PO
in the opposite direction CA' negative. It will be But PN=DN+DP. Hence,
found that by this arrangement, all trigonometrical

DN DP ME , DP quantities will change their signs upon passing

sin. (m+®')= +

PO
PO
PO

PO through zero and infinity. As the radius revolves

ME MO DP PM through the angle aCA' from Ca towards CA, the

+ sine diminishes, but is still positive, and the cosine

MO' PO PM' PO increases, and is negative. Thus, at Ca the cosine

DP EO
By the similar triangles,

Hence, passes through zero, and changes its sign. When

PM MO the radius coincides with CA' the sine : 0, and the

sin.(+')=sin.cos.«'+ sin.c' COS.c. [1.] cosine coinciding with CA' is = -1.

In like manner NO Through the angle A'Ca' the sine increases, and

OEF NE = OE FDM. the cosine diminishes, both being negative, the sine

OE-DM

Hence, cos.(w+w')= + changing its sign in passing through o at CA'; and

PO

D

A

M

P

sin.c=

PO

COS.w=

cos..'

+

PO

+

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COS.W-COS.co

.

-COS.

-COS.a

OE OM - DM PM

sin.c+sin.com tan. 1(+)
OM PO PM PO
sin.csin.co

[18].

tan, i comunale) DM ME By similar triangles,

Hence,

sin.c +sin.com
PM
MO

= tan. (+)

[19].

cosa +cos.ch cos.(m +') = cos.. cos.c' 7 sin.“ sin.'. [2].

sin.w+sin..! Throughout this investigation, the upper sign re

-cot.š(commande') [20]. fers to the first figure, and the lower to the second. (19.) From the formulæ [1] may be deduced a

sin.com sin.co
tan.com')

[21]. group of four others by merely adding, subtracting, cos.o+cos.co multiplying, and dividing them.

sin.csin.co

-cot. i{w ta') 10. By adding them we obtain

[22]

COS.O.--cos.c' sin.(w ta') + sin.( 62) = 2 sin.c cos.co' [3].

cos.ctcos.co' 2o. By subtracting,

-cot.f(w+6)cot. (comme') [23].

cos.cos.com sin. ( + )-sin.(1') = 2 sin.ce' cos.com

(24.) The sine and cosines of the doublet of an 30. By multiplying, sin.(-+-') sin.(-2)=sin

. o cos.'a'-sin.cos.en amaite may be found by making «="' in [1] and [2]. Eliminating cos. Po and cos. 200 the result is,

sin.2w=2sin.qcOS.. cos.26 = cos.*w-sin. *

w2cos.2.-1. sin.(w+') sin.(W—') = sin. *— since' [5]. 40. By dividing,

(25.) If in the last formulæ 2. be changed into a

and therefore w into do we find the values for the sin.(to sin.c cos.' *sin.a'cos.c

sine and cosine of half an angle. sin.(a) sin.w cos.c'sin.c' cos.co

sin.w=(1-COS..) cos. -=(1+cos.co). Eliminating the sines and cosines of w and e' the

(26.) The tangent of half an angle may be derived result is, sin.(to+w')_tan.cttan.c?

from these by (10) and (16), and hence we find,

[6]. sin.(com) tan.comtan..

sin.co

tan.wi= (20.) In a similar way the following group of

1+cos.co sin.com

1 + cos.co four formulæ may be deduced from [2] by addition, subtraction, multiplication, and division. cos.(w to + cos.(com') = 2 cos.ccos.co [8].

SECTION IV. cos.(+2) – cos.(-) =— 2 sin.c sin.c. [9].

On the Solution of Plane Triangles. cos.(+)cos.(c) = cos."+cos.l'-1 [10]. cos.(+6') tan.atan.a

(11].

(27.) In a plane triangle there are six parts, the cos.(c) tan.c'+tan.co

three sides and the three angles. In general, if the (21.) Given the tangent of two angles, to deter. magnitudes of any three of these six quantities be mine the tangents of their sum and difference. given, the magnitudes of the other three may be By dividing [1] by [2], we obtain,

computed by the aid of the formulæ and tables of sin.(+6') sin.ccos.o+sin.c'cos.co

trigonometry. cos.(wte) cos. cos a+sin. e sin.”

To guide us in some degree in the determination Eliminating the sines and cosines by the values of formulæ exhibiting the relations of the parts of for the tangents in (10), we find

a triangle, it is necessary that we should attend to

the nature of the tables, by the aid of which these tan..+tan.co tan.(-+) =

[13]. formulæ are to be computed when particular num1+tan.can.co

bers are substituted for their general symbols. (22.) From the formulæ [3], [4], [8], [9], we These tables are in general logarithmic, and by can derive the following group, by considering them, whenever an angle is known, the logarithms (+') and (cm') as single angles, and c and aias of its sine, cosine, tangent, &c. can be determined, their half sum and half difference, expressing the and, vice versa, when any of the latter are known, former by- and c', and consequently the latter by the angle can be found. Formulæ, therefore, in 1(+') and da').

order to be suited to such tables, should be such sin..+sin.'=2sin. ("+cos. (')

(14) as are adapted for logarithmic calculation, and sin.c_sin. '=2sin. :(w')cos. (w+w') [15]

therefore their different parts should be united as cos. +cos.«'=2cos. }(w+)cos. (c)

much as possible by multiplication, division, invo[16].

lution, and evolution; and as little as possible by cos.comcos.o'=-2sin. d(utc)sin. (cm) [17]. addition and subtraction.

(23.) From these four a group of six others may Further, as it would be impossible in any tables be immediately deduced by division. Let the first to give the values of the sine, cosine, &c. of angles be divided successively by the second, third and of all magnitudes, it must frequently happen that fourth; the second by the third and fourth; and the the angle, sine, or cosine, &c. which we seek, lies third by the fourth. After this division, the sines between two successive tabulated angles, sines, or and cosines of (+-') and (') being eliminated cosines, &c. We can in this case only compute by the formulæ found in (io) and (16), we obtain the true value approximately, and the degree of the the following formulæ:

approximation frequently depends on the formula Vol. XVIII. Part I.

a

2ab,

[ocr errors]

a

с

с

a

[ocr errors]

a

which we use. A formula which is proper when

sin. A

From the formula the angle is great, is often ill suited if the angle to

sin. B b be computed be small. Hence it is necessary in

sin. A + sin. B a+b

we deduce some cases to establish several different formulæ

sin. A sin. B ab' for the solution of the same problem, some being

a+b_tan.&(A+B) fitted for calculation in the cases where others fail, which by [18] becomes

a tan. (A-B) or give results deviating considerably from the

(30.) Let t be the segment of the side t between truth.

the perpendicular p and the angle C. By the defiThe Solution of Right-angled Triangles. nitions we have t= a cos. C. But also Cʻ = p + (28.) of the six parts of a triangle, it is only (6t) = p + 6° +2bs.

But a* = po + 1o ::

c* = a* + bo 2 ab cos. C. necessary to consider four in the case of a rightangled triangle; since the right angle is always

By adding the equations given, and one of the acute angles is the compli

c = a' + 622 abcos.C. ment of the other.

O= 2ab Let a and b be the sides, c the hypothenuse, and

we obtain A the angle opposite the side a.

c = (a + b) — 2ab(1 + cos.C), By (8.) we have

or c®=(-1) + 2ab( 1 cos.c).

Hence by [25] b

b =sin. A, =cos.A, :: tan.A, cot.A.;

4abcos.C= (a + b)-0, b

4absin."{C=-(

ab) + co. and by plane geometry, ao+b=c. These equations are sufficient for the solution of By multiplying and dividing these, observing the

equation all cases of right-angled triangles. All questions

sin. C=2sin. C cos. C, as to their solution must come under some of the

we obtain following four:

4a2basin.'C=[(a+b)/(?] [-(-6)* + c'], 10. Given the two sines, to find the hypothenuse and either angle.

(a + b) — C. c= a + b*, :lc = ila° +62)*,

All these results may be easily adapted for logarithtan. A .:: l.tan. A = la_ lb.

mic calculation. Let b

28 = a + b + C, 20. Given the hypothenuse and one side, to find the

2(8-a)=b toother side and either angle.

28-b) =a +-6, as co-= (c -6°, :;la = ilc+b) + Al(c-6),

2(8 — c)=b ta-C. b

Also sin.B ...l.sin. B = 16 lc,

(a+b) = (a + b + c)(a + bc),

-a-b) + c = (a +c-6)(c +b-a). b cos.A= ... lcos. A= lb lc.

Hence the four equations already obtained, when

divided by 4ab, give 30. Given the hypothenuse and one angle, to find the two sides.

8(8-0) cos. RC =

ab
b = c sin. B,

a= c cos B,
lc + lsin.B, la lc +lcos.B.

ab 4o. Given either side and an angle, to find the hy.

48.(8-a)(3-6)(8-0)

sin.?C= pothenuse and the other side.

b
b

b

b sin.B

(s-a)(sb) cos. A' tan.B cot. A'

tan.' C

8(s—c)
lb - 1 sin. B
- I cos. A,

which are all suited to logarithmic computation. la 1b -Itan. B 16 - Icot. A.

(30.) The formulæ which we have now establishThe Solution of Oblique Angled Triangles. ed are sufficient for the solution of oblique plane (29.) Let a, b, c, be the three sides, and A, B, C, triangles. The data in all such problems may be the angles opposed to them respectively. Let p be reduced to the following: the perpendicular from any angle A on the opposite

1o. Given two sides and the angle opposite to one side

of them.

20. Given two sides and the angle included by We have obviously sin.B

P
sin.C
р.

them.
b

3o. Given two angles and the side opposite to one sin.B 6 Elminating p, we obtain

of them. sin.C

4o. Given two angles and the side between them. sin.c

sin. A and, in like manner,

5o. Given the three sides. sin.A

sin. B 6. We shall consider these problems successively. Hence “ the sides of a plane triangle are as the 1°. Let a and b be the given sides, and A the sines of the opposite angles.”

given angle.

[ocr errors]

с

::16

sin.' C=(sa) (8-6)

aob?

a =

·lc

= 16

с

с

с

[merged small][ocr errors]

• The radius should be introduced when it is not = 1(as here supposed.)

a

It may

. a.

b sin.B sin. A, .: Isin.B= 16 + lsin. A-la.

(4.) All great circles bisect each other, and a se

condary bisects all parallels to its primary. By this formula, sin. B becomes known, but, except

1o. The intersection of the planes of two great in certain cases, the angle B will be equivocal. circles is necessarily a diameter of the sphere, and Since an angle and its supplement have the same a common diameter of both circles. Hence they sine, and neither necessarily exceeding two right necessarily bisect each other. angles, they may be each an angle of a triangle; 2o. By (3.) the plane of a secondary passes therefore we can only determine that B is either of through the axis of its primary, and therefore two angles which are known and supplemental, but through the centres of all parallels to the primary. it is in general impossible to decide which it is. Hence it bisects all such parallels. If b <a, the angle B must be acute.

(5.) The angle under two great circles is equal to therefore be found in this case.

the angle under their planes. If it happen to be known that b< c, though cit The intersection of the planes of two great circles self be not known, the same conclusion follows. is the common diameter joining the points of inter

The angle B being computed, the angle C be section of their circumferences. Tangents to the comes known, and thence the side c may be com circles drawn from these points of intersection are puted by

necessarily in the planes of the circles and persin.C.

pendicular to their common diameter; hence the sin. A

angle under these tangents is at the same time the 2°. To compute the angles it is only necessary angle under the circles, and the angle under their that the ratio of the sides be given. Let this ratio planes. be a : b. By (29) we have

(6.) The angle under two great circles is equal to - 6

the distance between their poles.
tan.(A-B)
tan. (A + B).

The axes of the great circles being perpendicular a + 6

to their planes are inclined at the same angle as the The value of A + B is found by subtracting C

planes of the circles. from 180°. Hence A-B becomes known by the

But the angle under the axes

is obviously measured by the arc which joins their preceding formula, and from this and A+B the

extremities, that is, by the distance between the values of A and B are derived. Having found A and B, C may be determined by poles of the great circles.

(7.) The angle under two great circles is measured sin.C sin.c=a

by the arc of a common secondary intercepted between sin. A

them. For this arc and the distance between the 3o. Let the angles be A, B, and the given side a. poles have a common complement.

sin. B Hence

sin.A
sin.(A+B)

SECTION II.
sin.A

of Spherical Triangles. The angle C is determined by C="-A-B.

(8.) Def. Three points upon the surface of a 40. Since the two given angles determine the sphere being connected by arcs of great circles, the third angle, this case is reduced to the last.

figure formed on the surface by these arcs is called 5o. This problem may be solved by any of the

a spherical triangle. four formulæ determined in (30.), and which are all

(9.) Any two sides of a spherical triangle taken suited to logarithms.

together are greater than the third side.

For if the radii of the sphere be drawn to the

three angles, they will form at the centre a solid SPHERICAL TRIGONOMETRY.

angle bounded by three plane angles which are SECTION 1.

equal to the sides of the proposed triangle. It is

proved (Eucl. lib. xi. prop. 20), that any two angles Of Circles on a Spherical Surface.

forming such a solid angle must be together greater (1.) By the principles of solid geometry it is

than the third. proved that if a sphere be intersected by a plane, (10.) The sum of the three sides of a spherical ihe section is a circle. If the plane pass through triangle is less than the circumference of a great circle. the centre of the sphere, the section is called a great For let any two of the sides a, b, be produced circle; if not, it is a lesser circle.

through the third side c until they meet again. (2.) That diameter of the sphere, which is per. The produced parts f--a and *-b will, with the pendicular to the plane of any circle of the sphere, third side c, form a triangle. Hence by the last is called the axis of that circle, and the extremities proposition, of this diameter are called the poles of the circle.

rata(3.) A great circle, whose plane is perpendicular

..:27 > a+b + c. to any circle, is said to be secondary to it. It is (11.) If the intersections of three great circles be evident that the planes of all secondaries to a circle the poles of three others, the intersections of the latter pass through its axis, and their circumferences pass will be the poles of the former. through its poles.

Let a, b, c, be three great circles, and a', b', c',

b=a

C= a

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axes.

2rA,

three others. The intersections of a and b are the four times the area of one of its great circles. poles of c', ; the intersection of the planes of a and Hence, h is the axis of c' and is perpendicular to every line

S=4r, in the plane of c'. Therefore it is perpendicular

•·L=2ra. to the intersection of the planes of a' and c'.

(19.) Cor. 1. Lunes of the same sphere are as Also, for the same reason, the intersection of the their angles. planes of b and c is perpendicular to the intersection (20.) Cor. 2. Lunes of different spheres are as of the planes of c' and a'. Since then the intersec- the products of their angles and the squares of their tion of the planes of cand a' is perpendicular to the intersection of the planes of b and c, and also to (20) Cor. 3. The whole surface of the sphere conthat of the planes b and a, it is perpendicular to two sists of four rectangular lunes. lines in the plane b, and is therefore perpendicular (22.) Cor. 4. A secondary to the sides of a lune to the plane b itself. Euc. lib. xi. Hence the in divides it into two equal rectangular isoceles triantersection of the planes a' and c' is the axis of b, gles, whose vertical angles are those of the lune. and the intersections of the circles « and c' are (23.) Given the surface of the sphere, to determine therefore the poles of b.

the area of a given triangle. In like manner it may be proved that the inter Let each pair of sides of the triangle be produced sections of b' and c' are the poles of a, and those of through the extremities of the third side until they a' and b' the poles of c.

intersect. There will thus be three lunes formed, (12.) If the poles of the sides of a spherical trian- whose angles will be the three angles of the triangle be joined by arcs of great circles, the sides of the gle, and their surfaces respectively will be triangle 80 found will be equal to the angles of the for

2r2B, 2rC, mer, and vice versa.

r being the radius of the sphere, and A, B, C, the Let a, b, c, be the sides of the first, and A, B, C, angles of the triangle (171.) the angles. By (6.) the sides of the second triangle

In the surfaces of these lunes the given triangle must be equal to the supplements of the angles of is three times repeated, and with it three of the the first, and by (11.) the sides of the first must be eight triangles into which the whole sphere is diequal to the supplements of the angles of the second. vided by the circles which form the given triangle.

Two triangles thus related are called polar tri. The surface of the hemisphere is equal to those angles.

three triangles together with the given one. Hence (14.) Def. Two semicircles being described the sum of the three lunes exceed the hemisphere upon the same diameter of a sphere, that part of the by twice the area of the given triangle. Theresurface of the sphere which they include is called a fore, if D be the area of the triangle lune:

D=ro(A + B + C)-pop, (15.). Def. The common diameter of the semi

..:D=rl(A + B + C)—.] circles called the axis of the lune.

Hence the area of a spherical triangle is equal to (16.) Def. The angle under the planes of the

that of a lune of the same sphere whose angle is semicircles is called the angle of the lune.

equal to half the excess of the sum of the three (17.) Lunes of the same sphere, whose angles are

angles of the triangle above two right angles. equal, have equal surfaces.

In the formula just obtained, the angles A, B, C, ,

B, F, For if their axes be supposed to be placed in co

are related to the radius unity. incidence as well as the planes of two of their If they should be expressed in seconds, it will semicircles, the planes of the other two semi. be necessary to divide each angle by 206265 (4.) circles being turned in the same direction, must (24.) Cor. 1. The areas of triangles on the sur. coincide, since the angles of the lunes are equal. face of the same sphere are proportional to the Therefore the semicircles which bound the lunes excess of the sums of their angles above two right will coincide, and therefore their surfaces will ne- angles. cessarily also coincide and be equal. (18.) Given the surface of the sphere and the angle

SECTION III. of a lune, to determine its area.

Let be the angle of the lune and S the surface Trigonometrical Formulæ cxpressing relations be. of the sphere. Let the angle c be divided into any tween the sides and Angles of a Spherical Trinumber of parts n. It is plain that the lune may angle. be divided into a number (n) of equal lunes; the angles of which will be Now as often as

(25.) The analytical formulæ, which expresses all is

the various trigonometrical relations of the sides contained in 360 degrees, so often will one of these and angles of a spherical triangle, although very lunes be contained in the whole surface of the sphere. numerous, and many of them apparently unconnectHence it appears that the lune whose angle is w, ed, may, nevertheless, be all derived from one forbears to the surface S the ratio w: 360°. If L be mula, which may be considered as the foundation the lune

of the whole structure of spherical trigonometry.

This formula, therefore, may be regarded in spheri. LES S.

cal trigonometry in the same point of view as that 360° 27

for the sine of the sum of two arcs in plane trigoThe surface of a sphere is proved to be equal to nometry, and, as in that case, we shall establish it

n

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