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BR : CR::LW (=AG +OD): wo (=cG+CD)

sine of the mean arc : cosine of the mean arc:: the (X)

sum of the sines of the extreme arcs : the sum of the

cosines of the extreme arcs. (Y) By comparing the above proportions, various other proportions may be formed. Thus BZ: BV::zx: zv by comparing (L. 119. with M.) that is double the sine of the mean arc : versed sine of double the mean arc::double the cosine of the mean arc : sine of double the mean arc, &c. for the rest.

(Z) By what has been already shewn, two arcs may be compared, and their sums and differences formed into a proportion. The arcs being in arithmetical progression, the

mean arc BE is half the sum of the extreme arcs Bo and BA. Therefore, in all the above proportions, and their variations, whereever the words mean arc occur, read half the sum of the arcs. Also the sine on, of the arc co, the common difference between the arcs, is the sine of half the difference between the arcs Bo and ba; hence in the above proportions for common difference between the arcs, read half the difference between the arcs. Thus for example (N. 119.) will run thus; radius : sine of half the sum of the arcs :: double the sine of half the difference of the arcs : difference between the cosines of the arcs.

(A) Hence it follows that the difference between the cosines of two arcs x radius=2 x sine of half the difference x sine of half the sum of these arcs.

But this has been obtained in another manner, see the fourth equation, F. 116.

Indeed from this proposition, the greater part of the formulæ noted in Prop. xvii. may be deduced, and it will be no bad exercise for the young student to trace out the coincidence,

PROPOSITION XX. (Plate I. Fig. 2.) (B) If three arcs be in arithmetical progression, the sum of the sines of the extreme arcs, is to their difference, as the tangent of the mean arc is to the tangent of the common difference between the arcs.

The same construction remaining as in the preceding proposition, AG+OD=sum of the sines of the extreme arcs BA and BO, and AP=AG-PG-AG-OD=the difference of the sines of the extreme arcs. But Ag+OD=AH, because th=DF=OD =PG.

By Prop. XIII. AH : AP::EK : EI.

Now Ek is the tangent of the mean arc BE, and Ei is the tangent of EO, the common difference between the arcs. Hence the proposition is evident. Q. E. D.

arcs.

: (C) Viz. If the extreme arcs be represented by A and B, of which a is greater than B, then sine A+sine B_tang (A + B)

(See Z. 120.) sine A-sine B tang I (A - B)

PROPOSITION XXI. (Plate I. Fig. 2.) (D) If three arcs be in arithmetical progression, radius is to double the cosine of the middle arc, as the sine of the common difference of the arcs, is to the difference between the sines of the extreme arcs.

Let the three arcs be BO, BE, and BA, as in the proceding propositions, and OE=AE the common difference between the

Then an is the sine of AE, and ap=difference between the sines of Ba and Bo (B. 120.)

The triangles Ain, cng, are equiangular and similar, for Ånc and ing are right angles; if therefore inc be taken away from both, there will remain the 2 ani = [cng;- but the triangle cng is equiangular with the triangle CEM.

Therefore ce : cm:: An: Ai; but Airip, for
AN : 10:: Ai : AP, and since an=lao, Ai=AP.
Hence CE : cm:: An: ip, and doubling the consequents,
CE: 2cm :: An : AP. Q.E.D.

(E) Viz. If A and B represent the extreme arcs, A being the greater. rad sine I (A-B)

(See Z. 120.) 2 cos (A + B) sine A-sine B

PROPOSITION XXII. (Plate I. Fig. 2.) (F) If three arcs be in arithmetical progression, the rectangle of radius, and the sum of the sines of the extreme arcs, is equal to the rectangle of double the cosine of the common difference of the arcs, and the sine of the mean arcs. The same construction remaining as in the foregoing propo

on X Ein sitions, CE: C:Em : ng=

But ag+OD=AH (B. 120.)=2ng.

For PG=GH (B. 120.) and ai=iP (D. 121.); now Ai +Gh= ip+Pg=ng, therefore ai +iP+PG+Gh=AH=2nQ.

2cm x Em Hence Ag + od =

that is (AG+OD). Ce=2cn . Em.

Q. E. D. (G) Let the extreme arcs be represented by A and B, then the mean arc will be } (A+B), and the common difference of

CE

CE

2 cos }(A-B). sine (A+D) and if rad=1, sine A+sine B= 2 cos 1 (A-B). sine (A+B).

Consequently. If the sine of the mean of three equidifferent arcs be multiplied by twice the cosine of the common difference, and the sine of either of the extreme arcs be deducted from the product, the remainder will be the sine of the other extreme arc ; the radius being 1.

COS A

OF THE SINES, COSINES, TANGENTS, &c. OF THE MULTIPLES

OF ARCS. *

(sine A . COS B)+(sine B. COS A) (H) Sine (A+)B=

(D. 115.)

rad Let B=2A, and rad=1, then we shall have Sine 3A=(sine A. cos 2A)+(sine 2A.cos A).

sine 2A But sine A - (O. 108.) and cos 2a=2 cose A-1

2 cos A (P. 108.)

sine 2A Therefore sine A . cos 2 A

::(2 cos’ A-1)=sine 2A.

2 cos A sine 2A

=(sine 2A.cos a) – sine A, and substitating this
2 cos A
in the first equation.

Sine 3A=(sine 2A.cos A) - sine A+(sine 2A . cos A).
Viz. sine 3a=2 (sine 2A . cos A)-sine A.
And by making B=3A, and following the same method,
Sine 4A=(2 cos A . sine 3A)-sine 2A. Hence we obtain
(1) + Sine A=sine A

Sine 2A = 2 cos A. sine A
Sine 3A=(2 cos. A . sine 2A) -sine A
Sine 4A =(2 cos A . sine 3A)-sine 2A
Sine 5A=(2 cos A . sine 4A)-sine 3A

Sine 6A=(2 cos A . sine 5A)-sine 4A, &c.
Where the law of continuation is evident.

OR, (K) By substituting the value of cos A=N1-sine A, &c. as in the single arcs (M. 104.), sine 2A found above, &c.

# Sine A=sine A.

Sine 21=2 sine A 1-sine’ A

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* See Euler's Introductio ad Analysin Infinitorum ; Traité de Trigonométrie, par M. Cagnoli; Crakelt's translation of Mauduit’s Trigonometry; Baron Masere's Trigonometry; Emerson, &c. + Cagnoli, page 27. Legendre, page 358.

Crakelt's translation of Mauduit, page 15.

Sine 3A=3 sine A-4 sine As
Sine 4A=(4 sine A-8 sine AV1-sine A
Sine 5A=5 sine A-20 sine3 A +16 sines A

Sine 6A=(6 sine A-32 sines A +32 sine: a) v1 - sine’A, &c:

OR, (L) Sine A=Ni-Cos? A

Sine 2A-2 COS AW1-cos? A
Sine 3A=(4 cos A-1)71 - cos? A
Sine 4A=(8 cos A-4 cos a) vi-cosA
Sine 5A=(16 cos4 A-12 cos' A+1) 71-cosA

Sine 6A=(32 cos54 — 32 cos'A+6 cosa) v1 -cos A,&c. (M) Cosine (A+B)= (COSB.COS A)-(sine A.sine B)

(D.115.)

rad Now if B=2A, and radius=1 as before, we shall obtain Cosine 3A=(cos 2A. Cos A)-(sine A.sine 2A) But sine A. sine 2A = sine? A2. Cos A (O. 108.)= (1 - cos 2A).COSA. (P.108.)=cos A-(cosa. cos 2A) therefore

Cos 3A=(cos 2A. Cos A) -cos A +(cos A.cos 2A)=(2 cos A.cos 2A) -cos A.

And by making B=3A, and pursuing the same method, we shall find cos 4A = 2 COS A.cos 3A -cos 2A. (N) *Cos A=cos A

Cos 2A=2 cos A.COS A-1
Cos 3A=2 cos A.cos 2A -COS A
Cos 4A=2 cos A.cos 3A -cos 2A
Cos 5A = 2 cos A.cos 4A-COS 3A
Cos 6A=2 cos A.cos 5A - cos 4A, &c.

OR,
(0) By substituting the value of cos as V1-sine’A,

ví cos 2A, found above, &c.

+Cos A=N1-sine? A
Cos 2a=1-2 sine A
Cos 3A=(1 — 4 sine? A)71 -sine’ A
Cos 40=8 sine A-8 sine A +1
Cos 5A=(16 sinet A-12 sine A+1) 71-sine' A
Cos 6A=i-18 sine A+48 sine A - 32 sine A, &c.

OR,

(P) Cos A=COS A

* Cagnoli, page 27. Legendre, page 358.

by sub

Cos 2A=2 cos A-1
Cos 3A=4 cos3 A-3 cos A-
Cos 4A=8 cos + A-8 cos? A +1
Cos 5A=16 cos5 A - 20 cos3 A +5 COS A
Cos 6A=32 cos 6 A - 48 cos* A +18 cos? A-1, &c.

tang A . tang B (Q) Tang (A+B)=

(C. 115. or K. 118.) rado-(tangA.tang B)

2 tang A Now if B=2A, and rad=1; tang 2A=

1-tanga

3 tang A - tang: A stitution, and tang 3 A=

Hence,
1-3 tang A
Tang A=tang A

2 tang A
Tang 2A=

1-tango A

3 tang A- tang3 A Tang 3A =

1-3 tang A

4 tang A-4 tang: A Tang 4A =

1-6 tango A+tang* A

5 tang A-10 tang: A+tang' A
Tang 5A=

&c.
1-10 tango A+5 tang*A
rad

1 * (R) And since tang A=

be substituted for

cot A cot A tang 1, we shall obtain Cot A=cot A

cot? A-1 Cot 2A=

2 cot A

cot A - 3 cot A Cot 3A=

3 cot? A-1

coto A-6 cot? A +1 Cot 4A=

-4 cot3 A-4 cot A

cots A - 10 cots A +5 cot A Cot 5A=

&c. 5 cott A-10 cot? A +1

i be substituted for cos A (N. 123.) the secant of Also if

sec A the multiple of any arc may be obtained.

; if

OF THE SINES AND COSINES OF THE POWERS OF ARCS.+ (S) Cos 2A=1—2 sine? A (O. 123.)

Cagnoli, page 29.

Crakelt's translation of Mauduit, page 34. + Cagnoli, page 30. Crakelt's translation of Maudụit, page 20.

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