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sin A + sin B = 2 sin 1 (A+B) cos } (A - B), (9)

} B, sin A – sin B = 2 cos } (A+B) sin } (A - B), (10) cos A + cos B = 2 cos } (A+B) così (A - B), (11) cos A - cos B =- 2 sin}(A+B) sin}(A-B). (12)

As the equations (5), (6), (7), (8) are perfectly general, that is true for all values of A and B, so the equation

sin A + sin B = 2 sin } (A' + B) cos } (A' B'), derived from them is perfectly general, i.e. true for all values of A' and B, and therefore for A and B. It is for this reason that we are at liberty to remove the accents after transformation. If in equation (1) we suppose A = B, we obtain sin (A + A) = sin A cos A + cos A sin A, sin 2 A = 2 sin A cos A. 2

(13) In the same way from equation (2) we obtain

cos? A - sin? A. If in this equation we substitute for cos’A its equal, 1 - sin’A, we obtain cos 2 A = 1

2 sin’A; if in the same equation we substitute for sin’ A its equal 1 - cos? A, we obtain

2 cos? A - 1; from which we have the following values for cos 2A: cos 2 A = cos A - sin’A,

(14) cos 2 A = I - 2 sino A,

(15) = 2 cos? A - 1.

(16) If we divide equation (9) by equation (10), we ob


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2A COS 2 A

If we ad

sin A + sin B sin A - sin B

2 sin } (A+B) cos }(A - B)
2 cos 1 (A + B) sin (A - B)'

we obta

sin A + sin B

sin Asin B

tan? (A + B) cot } (A - B).

by subt


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If on th 24 by right-h

Substituting for cot }(A - B) its value, the recipro-
cal of tan } (A - B) Chap. I. (7),
sin A + sin B tan (A + B)

sin A - sin B tan(A - B)'
If we divide equation (1) by (2), we obtain

sin A cos B + cos A sin B tan (A + B) =

cos A cos B sin A sin B Dividing the numerator and denominator of the right-hand side by cos A cos B,

sin A sin B


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cos B

1. Rela


1. -In the s

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c res


tan (A + B)

sin A sin B

cos A' cos B Therefore

tan A + tan B tan (A + B)


tan A tan B In a similar manner from (3) and (4) we obtain

tan A - tan B tan (A - B)


I + tan A. tan B If in (18) we make A = B, we obtain

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ang dicu by hav



2 tan A

tan A


tan 2 A


If we add together the equations

1 = cos’ A + sin’A, COS 2 A = cos A - sin’A,

I =

we obtain




1 + cos 2 A = 2 cosA;

(21) by subtraction we obtain

2 sin’A. If on the left-hand side of these equations we replace 2 A by A, we must write A instead of A, on the right-hand side; therefore 1 + cos A = 2 cos LA,

(23) = 2 sino 1 A.






1. Relation between the sides and Angles.--2. Expression for the

Area.-3. Five cases of Plane Triangles. I. Relation between the sides and angles. -In the triangle ABC (fig. 12), let the lengths of the sides BC, AC, and AB be denoted by a, b, and c respectively, and let the values of the opposite angles be A, B, and C. From C draw CP

perpendicular to the side AB, and let its length be denoted by p; then from the right-angled triangle ACP we have (Chap. III. Prop. I.)

p = b sin A;

P from the right-angled triangle BCP,

p =

= a sin B;


sin B

b sin A,

a:6:: sin A : sin B. From this follows



The sides of a triangle are in the same ratio as the sines of the opposite angles.

This proposition remains true if one of the angles be oblique, as in fig. 13. For in the triangle BCP, p-- a sin B; and in the triangle ACP, p = b sin PAC = b sin A ; because, since the angles A and PAC are supplemental, their sines are equal (Chap. II. Sect. 3), and therefore, as before,

a sin B = b sin A. From the proportion

a:b:: sin A:sin B, it follows that

a + b: a b::sin A + sin B:sin A - sin B. By equation (17) Chap. IV., sin A + sin B : sin B - sin B

:: tan ] (A + B): tan ] (A - B), therefore a + b: a - b :: tan } (A + B) : tan} (A - B). +

: - B From this follows:



In a triangle the sum of the sides is to their difference, in the same ratio, as the tangent of half the sum

of the base angles, is to the tangent of half their difference.

All the relations which exist between the sides and angles of a triangle may be derived from the expression, which gives the value of the cosine of any angle in terms of the sides. This expression may be deduced as follows:-in fig. 12 let the angle A be acute, we have (Euclid, Book II. Prop. XIII.)

BC2 = AC2. + AB? – 2AB ~ AP, but AP = AC.cos A. (Chap. III. Prop. 1.) Substituting this value for AP, and writing a.b and c, for BC, AC, and AB respectively, we obtain

a= 12 + O2 2bc cos A; solving this equation for cos A,

62 + c2 a?

2bc Similar values may be obtained for cos B and cos A, which together form the group of formulæ,

12 + c2-a

c2 + a2 - 62

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cos B


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cos C

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a> +62 - c

2ab If the angle A be oblique, as in fig. 13, we obtain the same expression for cos A, for (Euclid, Book II. Prop. XII.)

BC2 AC + AB2 + 2AB ~ AP, but AP AC cos PAC = - AC cos A,

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