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But the angle C is equal to the difference between the angles a and b (Geom. B. I., P. 25, C. 6): hence,

sin (a - b) = sin a cos 6 cos a sin b; . (a) that is, The sine of the difference of any two arcs or angles is equal to the sine of the first into the cosine of the second, minus the cosine of the first into the sine of the second.

It is plain that the formula is equally true in whichever quadrant the vertex of the angle C be placed: hence, the formula is true for all values, of the arcs a and b.

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72. To find the formula for the sine of the sum of two angle

or arcs.

By formula (a)

sin (a - b) = sin a cos b cos a sin b, substituting for b, b, and recollecting (Art. 66) that,

sin (-2)

sin a and cos (-x) = cos x; and also that a-(-1) = a + b, we shall have, after making the substitutions and combining the algebraic signs,

sin (a + ) = sin a cos b + cos a sin b. (6)

73. To find the formula for the cosine of the sum of two

angles or arcs. By formula (6) we have,

sin (a + b) = sin a cos b + cos a sin b, substitute for a, 90° + a, and we have, sin [(90° + a) + b] =sin (90° + a) cos b + cos (90° + a) sin b.

But, sin [90° + (a + b)] = cos (a + b) (Table II.) :

sin (90° + a) = cos a, and,

cos (90° + a), = – sin a ; making the substitutions, we have, cos (a + b) =

= cos a cos / sin a sin b. (c)

74. To find the formula for the cosine of the difference between

two angles or arcs.

By formula (6) we have,

sin (a + b) = sin a cos b + cos a sin b. For a substitute 90° — a, and we have, sin [90° — (a - b)] = sin (90° — a) cos b + cos (90° — a) sin b. But, sin [90° — (a - b)] = cos (a - b) (Table II.),

sin (90° – a) = COS ag

cos (90° – a) making the substitutions, we have,

cos (a - b) = cos a cos 6 + sin a sin b. . (d)

sin a;

75. To find the formula for the tangent of the sum of two

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COS a COS

dividing both numerator and denominator by cos a cos b, sin a cos 3

COS a sin 7

+
= cos a cos b

22
sin a sin b
1

cos a cos 7'

tan a + tan 6 tan (a + 1) =

1

tan a tan 6

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76. To find the tangent of the difference of two arcs.
tan (a - b) =

sin (a - b), (Table 1).

cos (a - b)
sin a cos 6
COS d sin 6

by (a) and (a). cos a cos b + sin a sin b Dividing both numerator and denominator by cos a cos 0,

tan a

tan 6
1+tan a tan 7

(9)

tan (a 0)

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77. The student will find no difficulty in deducing the following formulas.

cot a cot b 1 cot (a + b)

(h)
cot a t cot b

cot a cot b + 1
cot (a
b)

(i)
cot b

cot a

.

78. To find the sine of twice an arc, in functions of the are

and radius.
By formula ()
sin (a + b)

= sin a cos b + cos a sin b. Make a = b, and the formula becomes,

sin 2a= 2 sin a cos a.

a

If we substitute for a, , we have,

2
sin a= 2 sin ja cos ja.

(k 1)

79. To find the cosine of twice an are in functions of the art

and radius, By formula (@)

cos (a + b) = cos a cos b - sin a sin b. Make a = - b, and we have, = cos? a

(1) By Table I., sino a = 1 - cos' a; hence, by substitution,

2 cos' a – 1. . (21) Again, since cos' a = 1 - sino a, we also have,

(12)

cos 2a

sin? a.

cos 2a

.

1 - 2 sin? a.

cos 2a

(m)

tan? a

80. To determine the tangent of twice or thrice a given arc in

functions of the arc and radius. By formula (f)

tan a + tan 7 tan (a + b)

1- tan a tan b Make b = a, and we have,

2 tan a tan 2a

1 Making b = 2a, we have,

tan a + tan 2a tan 3a

1

tan a tan za i substituting the value of tan 2a, and reducing, we have,

3 tan a tan 3a =

;

(m 1)

1 3 tan” a The student will readily find

cot a

tan a cot 2a

(n) 2

tan a

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81. To find the sine of half an arc in terms of the functions

of the arc and radius. By formula (1 2)

cos 2a = 1

2 sin? a. For a, substitute sa, and we have,

:1

2 sino ja; hence,

2 sino ja

1

1 sin ja

(0) 2

COS (

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COS at

82. To find the cosine of half a given are in terms of the

functions of the arc and radius. By formula (1)

2 cos? a – 1. For a, substitute ja, and we have,

cos a = 2 cos? ja – 1;

COS 2a

hence,

cos ja

1 + cos a

2

(p)

COS a

83. To find the tangent of half a given arc, in junctions of

the arc and radius. Divide formula (0) by (p), and we have,

1 tan ja

(9)

1 + cos a Multiplying both terms of the second member by V1 - cos a,

1 COS a and reducing tan sa

(21)

sin a Multiplying both terms by the denominator V1 + cos a, and reducing tan ja

(q 2) 1 + cos a

sin a

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GENERAL FORMULAS.

84. The formulas of Articles 71, 72, 73, 74, furnish a great number of consequences; among which it will be enough to mention those of most frequent use. By adding and subtracting we obtain the four which follow,

sin (a + b) + sin (a - 1) = 2 sin a cos b, (9) sin (a + ) - sin (a )

) = 2 sin b cos a, (s) cos (a + 1) + cos (a ) ) 2 cos a cos b, (1)

cos (a 1) – cos (a + 1) 2 sin a sin b, (2) and which serve to change a product of several sines or cosines into linear sines or cosines, that is, into sines and cosines multiplied only by constant quantities.

85. If in these formulas we put a +b = p, a b = 9, which gives a =

1 + 2

0
P р 9

we shall find
2
sin p + sin q = 2 sin ( 2 + 1) cos } ( 2 1),. • (2)

sin q = 2 sin }(P 2) cos }(P + 9), . . (c) cos P + cos q = 2 cos }(p+1) cos } (p ),

(3) ios q - cos p= 2 sin }(P + 1) sin 3 (P - 1), .

sin 1

.

.

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