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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, COS 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 triangle.


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

sin (A+B). sin (A — B) = sin2 A

sin (B+c). sin (B-c) = sing B

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sin2 B,

sin2 c,

sin (c + D). sin (C–D) = sin2 c sin D,

&c. &c. &c.

sin (LA) sin (LA) = sin2 L - sin2 A.

L, be

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 (L+A) + sin (L− a) = 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 (L+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 circumferences; 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 5D' + &c = 0,

COS D' + COS 3D' + COS 5D' &C=0;} (XXV.)

cos + &c 0.

If E' were equal 2D', these equations would become sin D ́+ sin(D'+E') + sin (D' + 2E') + sin (D'+3E′) + &c=0, COS D'+cos (D'+E') + cos (D'+2E') + cos (D'+3E)+&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 ratio of 2 to 1. To investigate a general expression for an infinite series of this kind, let

ssin 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 1-2z. cos B + z2, making

2 = 1.


sin A +z [sin (A + B) — 2 sin ▲ COS B] 1-2%. COS B +22

Now this fraction will be =
Therefore, when z = 1, we have


sin A+ sin (A + B) — 2 sin A. COS B

COS B = Sin (A


2-2 cos B

; and this, because 2 sin A. + B) + Sin (A — B). (art. 21), is equal to sin ▲ — sin (^—B). But, since sin A'- sin B'= 2 cos(A'+B'). 2(1- -COS B) sin (A-B'), by art. 25, it follows, that sin a-sin (A—B)= 2 cos (AB) sin B; besides which, we have 1 COS B 2 sin B. Consequently the preceding expression becomes s = sin 4 + sin (A + B) + sin (A+2B) + sin (▲+3B) + &c, COS (A-1B) ad infinitum =

2 sin B

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35. To find the sum of n + 1 terms of this series, we have simply to consider that the sum of the terms past the (n+1)th, that is, the sum of sin [A+(n+1)B] + sin [A+(n+2)B] + sin [A+ (n + 3)] + &c, ad infinitum, is, by the preceding cos [A+ (n+B] theorem,= Deducting this, therefore, from the former expression, there will remain, sin a + sin (A+B) + sin (A + 2B) + sin (A + 3B) + . . . . sin (A + NB) = COS (AB))—cos [A + (n + 1)B] sin (A+B). sin (n+1)B. (XXVII.)

2 sin B

2 sin B

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sin B

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


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COS ACOS (A + B) + COS (A+2B) + COS (A + 3B)+....

..... (cos A+NB) = (COS

COS (A+B). Sin (n+ I)в

sin B

.... (XXIX.)

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 Trigonometry, though it has been long known to mathematicians. Let the three arcs A, B, C, together make up the whole circumference, O: then, since tan (A+B) = R2 (tan A+ tan в) (by equa. xxIII), we have R2x (tan A+tan B+ tan c) = R2x [tan A+ tan в-tan (A+B)]=R2 × (tan A + )=, by actual multiplication and re

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art by sinc; then adding together the equamansioned, and reducing; there will result,

---4)+ sin c.sin (A−B)=0, c. sin (A—B)=0. Crese wentacions Craning for any three angles whatFer pur enemy tree angles of any triangle.

14. Le de seres frogs A, B, C, D.... L, be

ZMETTLER der ve ve 24

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3 = ese entons de mied gerber, the second memThe quot vil mist, and of consequence we shall

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ST 1-4=0.

Ting II i smir manner with sin (1-8), cos (A+B), 3-$, there will at length be obtained


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1. form an arithmetical Tid he is term is 0, the common differNyumber of circumferences; 2.0-3=0.5A+B=D, B+c=3D', the preceding equa

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Fr. Shese agations would become
-=++ sin (D'+3E) +&c=0,
-282-2)+ cos(D'+ SE) + &c=0

Version however, only shows the s

MAY 2 As a regles in arithmetical
Curence is to the first ter
Ce este a general expressi


rss recurring series, w Acase from the dev

TORAN is 1-2.cos B


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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 those tangents.


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 therefore apply to the angles of a plane triangle, which are, together, measured by arcs constituting a semicircle. So that, if radius be considered as unity, we shall find that, the sum of the tangents of the three angles of any plane triangle, is equal to the continued product of those tangents, (XXXI.)

37. Having thus given the chief properties of the sines, tangents, &c, of arcs, their sines, products, and powers, we shall merely subjoin investigations of theorems for the 2d and 3d cases in the solutions of plane triangles. Thus, with respect to the second case, where two sides and their included angle are given:

By equa Iv,

By compos

and division

ab sin A: sin B.

a+b: a- b sin A + sin B: sin A →


sin B.

But, eq. XXII, tan¦(A+B); tan (A — B) : : sin a + sin B : sin A-sin B; whence, ex equal. a+b: a-b: tan (A+B); tan (AB). . . . (XXXIÏ.)

Agreeing with the result of the geometrical investigation, at pa. 10, vol. ii.

38. If, instead of having the two sides a, b, given, we know their logarithms, as frequently happens in geodesic operations, tan (AB) may be readily determined without first finding the number corresponding to the logs. of a and b. For if a and b were considered as the sides of a right-angled triangle, in which denotes the angle opposite the side a, then would tan. Now, since a is supposed greater than b, this angle will be greater than half a right angle, or it will be measured by an arc greater than of the circumfer ence, or than O. Then, because tan (-)=1+ tan tan o tan 4-tan 10, and because tan OR = 1, we have


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tan (? — § 0) = ( − 1 ) ÷ (1 + † ) = And, from the preceding article,

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a-b tan (A-B) tan (A-B)


a+b tan (A + B)


tan (A-B) cot c. tan (90) ... (XXXIII.)

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