Page images
PDF
EPUB

ference of the sines; it follows, that the first of these equations converted into an analogy, becomes sin (A-B): sin A-sin B ; : sin at sin B : sin (A + B)(XX.) That is to say, the sine of the difference of any iwo arce or angles, is to the difference of their sincs, as the sum of thase sines is to the sine of their sum.

If A and B be to each other as n + 1 to n, then the preceding proportion will be converted into sin a : sin (n + 1) A sin na :: sin (n + 1)^ + siu na : sin (2n + 1) A (XXI.)

These two proportions are highly useful in computing a table of sines: as will be shown in the practical examples at the end of this chapter.

25. Let us suppose A+B=A', and a B = B'; then the half sum and ihe half difference of these equations will give respect vely A= ('+B'), and B = {(A - B'). Putting these values of A and B iu the expressions of sina COSB, sin B . cos A, COS A. COS B, sin À si: B, obtained in arts. 21, 22, 23, there would arise the following formulæ :

sin į (A' +B). cos SA - B R(sin A' +sin B'),
sinį (A-B'). cos jíd'+B Risin a'-sin B),
cos} ('+B').cos la – B') R(COS A'+cos B),

sinį (A+B'). sin ia'-B') - R(cos B - COS A'). Dividing the second of these formulæ by the first, there will be had sin}(A'B') cos }(A+B')_sin (A' – B') cos(^'+b') sina'-sinB' sinsk A'+B') cos(A-B') C: {(A' B') sin(A' + B') sin A'+sing'

sin But since

it follows that the two factors of the first member of this equation, are

respectively; so that the equation tani+B)

tan (A-B') sin A' - sin B' manifestly becomes tan (A + B')

(XXII.)

sin a'+sin B This equation is readily converted into a very useful proportion, viz. The sum of the sines of two arcs or angles, is to Their difference, as the tangent of half the sum of those arcs or angles, is to the tangent of half their defference.

26. Operating with the third and fourth formulæ of the preceding article, as we have already done with the first and second, we shall obtain tan | (A' + B') tan (A - B')

cos A'

COS A' + cos B In like manner, we have by division, sina' tsinB _ sin}(A+B')

=tan(A' +B');

sina' + sin B' cosa'+cos B cos)(A'+3')

= coti(A-B'); sin A'

sin A'. =tan(A' B')... COSA +COSB

cot {(A' + B').

COS

tan

[ocr errors]

R

i and

COS

R

[ocr errors]

tan

tan (' -B), and

R

R

COS B

R2

COSB - CO A

-IB

SIA B

COS B-COSA

COS B

COSA

.

cos A' + cos B' cot ) (A' + B')

tan (A B') Making B = 0, in one or other of these expressions, there results, sin A'

1
= tan JA'.
1+
OSA

cot SA'
sin A'
= cot ja'

( xxii.) 1

tan ja" 1 + cos A co SA

1

=c01' fa' = 1- cos A tan sa

tan 2 A These theorems will find their application in some of the investigations of spherical trigonometry.

COSA

[ocr errors]

:

COS A

sin B

tan

COS

R

27 Once more, dividing the expression for sin (A + B) by that for cos (A B), there results sin (A + B)

sin A cos e + sin B .cos A C S(A + B)

COS B. I sin A then dividing both numerator and denominator of the second

sin fraction, by cos A . cos B, and recollecting that

we shall thus obtain tan (A + B) R ('an a + tan B)

RIF tan A

R' (an a + tan B) or, lastly, tan (A + B) =

(XXIII.)

R2 F tan Alan B
Also, since cot = we shall have

tan
R?

R? 7 tan A . tan
cot (A + B)=

tan (A + B) tan A + tan B which, after a little reduction, becomes

cota, cot B F R % cot (A + B)

(XXIV.) cot B + cot A

R

tan B

.

R2

[ocr errors]

=

28. We might now proceed to deduce expressions for the tangents, cotangents, secants, &c. of multiple arcs, as well as some of the usual formulæ of verification in the construction of tables, such as sin(540+^) + sir(54- A)-sin(180+ A)-sin(18°_A)=sin(90°-1); sina + sin(36"-A)+sin(720 + A)=sin(36° + A) + sin(72o - A).

&c. &c. But, as these enquiries would extend this chapter to too great a length, we shall pass them by ; and merely investigate a few properties where more than two arcs or angles are concerned, and which may be of use in some subsequent part of this volume.

29. Let

COSC

A.

29. Let A, B, C, be in any three arcs or angles, and suppose radius to be unity; then

sin A . sin c + sin B. sin (B + c) =

sin (A + B + c).

sin(A + B) For, by equa. v, sin (A+B+C) = sin A . cos (B+c) + cos A sin (B+C), which, (putting cos B cos C - sin B . sin c for cos (B + c)), is = sin A . COS B. - sin sin B . sin c' + COS A . sin (B +c); and, multiplying by sin B, and adding sin A . sin c, there results sin A . sin c + sin B . sin (A+B+c) = sip A. COS B , cos C sin B + sin A . sin c . cos? B + COS A . sin B. sin (B+C)=sin A .COS B . (sin B .cos c+cos B. + cos a . sin B. sin (B+c) = (sin A.cosB + cos a . sin B)* sin (B+C) = sin (A + B). sin (B +.c). Consequently, by dividing by sin (A + B), we obtain the expression above given.

sin c)

In a similar manner it may be shown, that
sin (B – c = sin A sin c = sin B sin (A – B+c).

sin (A - B) 30 If A, B, C, D, represent four arcs or angles, then writo ing c+d for c in the preceding investigation, there will result, sin (B+c+=

sin a.sin(c+p)+sing. sin(a+b+c+d),

sin (A + B) A like process for five arcs or angles will give sin(B+c+D+E)= sin a sin (c+d+8) +sias. sin(a+b+c+o+e).

(A + B) And for any number, A, B, C, &c. to L,

sinA.sin(c+D+...L)+sins.sin(A+B+c+ ...) sin (B+6+ ....L)=

sin (A + B) 31. Taking again the three A, B, C, we have

sin (B-) = sin B . cos C-sin c , cos B,
sin (C-A) = sin c . cos A-sin a cos C,
sin (A - B)

sin A .COS B-sin B . CoS A. Multiplying the first of these equations by sin A, the second by sin B, the third by sin c; then adding together the equations thus transformed, and reducing ; there will result, sin a : sin (B -c)+sin B . sin (c-A) + sin c

sin (A-B) =0, ços A . sin (B-c)+cos B .sin(c-A) + cos c. sin (A-B)=0.

These two equations obtaining for any three angles whatever, apply evidently to the three angles of any trianglea

32. Let the series of arcs or angles A, B, C D contemplated, then we have (art. 24),

L, be

sin

[ocr errors]

sin (A + B). sin (A B) = sin? A – sino B,
sin (B + c) sin (B-C) sino B -- sinac,
sin (c + D). sin (c-D)

= sinc-
&c. &c. &c.

sin 2 D,

sin (L + A). sin (L-A) = sin? 1- sino A. If all these equations be added together, the second member of the equation will vanish, and of consequence we shall have sin (A + B) sin (A--B) + sin (B+c) sin(B-c) + &c...

+ sin (1+) + sin(L--6)=0. Proceeding in a similar manner with sin (A - B), cos (A + B), sin (B -c), cos (B+c), &c. there will at length be obtained cos (A+B) sin( A - B)+cos (B+c). sin (B -c) + &c...

+ cos(i+a). sin (L-A) = 0. 33. If the arcs A, B, C, &c. ... L form an arithmetical progression, of which the first term is 0, the common difference d', and the last term L any number n of circumterences ; then will B-A=D', c-B =D', &c. A + B =D', B +c= 3D', &c : and dividing the whole by sin d', the preceding equations will become sin d' + sin 3D + sin 5p' + &c: 0,2

:0;} cos D' + cos 3D' +cos 50' + &c.

(XXV.) If e' were equal 2D', these equations would become sin d'+ sin (D'+E')+sin (D+22') + sin (D'+3E')+ &c=0, cos D'+ cos (d'+e') + cos (0'+22') + cos (D'+38')+ &c = = 0.

34. The last equation, however, only shows the sums of sines and cosines of arcs or angles in arithmetical progression, when the common difference is to the first term in the ration of 2 to !: To investigate a general expression for an infinite series of this kind, let

s + sin a + sin (A + B) + sin (A + 2B) sin (A +3B) + &c. Then, since this series is a recurring series, whose scale of relation is 2 cos B – 1, it will arise from the developement of a fraction whose denominator is ! 2z.cos a + z2, making z=1. Now this fraction will be =

sina += [sin(A + B)-2 sina.cos B).

1-25. COS B + 22 Therefore, when z = 1, we have SIN A + sin (A + B)-2 sm A . cos B

'; and this because 2 sin A., 2-2 cos B cos B = sin(A + B) + sin (A-B) (art. 21), is equal to sina - Sir (A-B)

But, since sin a-sin b' = 2 cos (A + B'). 2(1-сos B)

[ocr errors]

sin lia!--B/), by art. 25, it follows, that sin A -sin (AB)= 2 cos (A -}e)sin B ; besides which, we have 1 - cos B = 2 sin? {B. Consequently the preceding expression becomes $ = sia a *sin (A + n) -+- sin (A + 2B) +- sin (A + 3) +- &c. ad infinitum = cos (A - B)

(XXVI.) 2 sin 3 35. To find the sum of n + 1 terms of this series, we have simply to consider that the sun of the terms past the (n + 1)th, that is, the sum of sin [1 + (n + 1)B} + sin(A +in+2jB1+ sin (1 + (n + 3)B1+ &c. ad infinitum, is, by the preceding

cos (A+<+ )B). Deducting this, therefore, from theorem,

2 sin B the former expression, there will remain, sin A + sin (A + B) + sin (A + 2B) + sin (A + SB) +. sin (A + NB) = cos(A - R)-[A+in+B] _sin(1 + B )sin(n+1 B

(XXVII.) 2 sin B

sin je By like means it will be found, that the sums of the cosines of arcs or angles in arithmetical progression will be COS A + cos (A + B) + cos (A + 2B) + cos (A + 3B) + &C«

sin (A - B) ad infinitum

(XXVIII.) 2 sin B

Also, cos A + cos (A + B) + cos (A + 2B) +cos (A + 3B) to.... (CO3 A + 1B) cos (A + B). sin }(n+1,B

(XXIX.)

sin B 36. With regard to the tangents of more than two arcs, the following property (the only one we shall here deduce) is a very curious one, which has not yet been inserted in works of Trigonon.etry, though it has been long known 10 mathematicians. Let the three arcs A, B, C, togсther make up the whole circumference, 0 : then since tan (A + B) = R? (tan A + tan B)

-(by equa.xxIII), we have R* *(tan AtlanB+ can C) = R$ * [tan A + tan B- tan (A + B)]=R® X (tan At

R? (tan Atlan B) tan B

), by actual multiplication and reduction, lo tana. tan c, since tan c tan [o

R(an A + tan B) (A + B)] tan (A - B) =

by what has preceded in this article. The result therefore is, that the sum of the tangents of any three arcs which together constitute a circle, multiplied by the square of the radius, is equal to the product of these langents. ..(XXX.)

Since both arcs in the second and fourth quadrants have their tangents considered negative, the above property will apply to arcs any way trisecting a semicircle ; and it will there. VOL. II.

fore

2 R

tan A. tan B

[ocr errors]

tan Atan B

tan B

R2-tan Atan B

« PreviousContinue »