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sin. a cos. b-cos, a sin. 6
sin. (ab)=

(3)
R
cos. a cos. b+sin. a sin. 6
cos. (ab)=

(4)

R (73.) Expressions for the sine and cosine of a double arc.

If, in the formulas of the preceding article, we make b=a the first and second will become

2 sin, a cos. a sin. 2a=

R

cos. 'a--sin. 'a cos. 2a=

R Making radius equal to unity, and substituting the values of sin. a, cos. a. &c., from Art. 28, we obtain

2 tang. a sin. 2as

1+tang. 'a

1-tang. 'a cos. 2a=

1+tang. 'a (74.) Expressions for the sine and cosine of half a given

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arc.

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COS, a=

If we put ja for a in the preceding equations, we obtain

2 sin. ja cos. ļa
sin, a=

R
cos. *}a-sin. ?{a

R
We may also find the sine and cosine of ja in terms of a.

Since the sum of the squares of the sine and cosine is equal to the square of radius, we have

cos. *}a+sin. 'Ja=R'. And, from the preceding equation,

cos. '}a-sin. 'la=R cos. a. If we subtract one of these from the other, we have

2 sin. 'la=R-R cos. a. And, adding the same equations,

2 cos. 'fa=R’+R cos. a. Hence, sin. ja= {RP-R cos. a;

cos. ja= VR'+IR cos. a. (75.) Expressions for the products of sines and cosines. By adding and subtracting the formulas of Art. 72, we obtain

COS.

2
sin. (a+b)+sin. (ab)=sin. a cos. 6.

R

2
sin. (a+b)--sin. (a--b)= cos, a sin. b;

R

2
(a+b)+cos. (ab)=r cos. a cos. b;

2 cos. (ab)-cos. (a+b)=í sin. a sin. b. If, in these formulas, we make a+b=A, and a-b=B; that is, a=3(A+B), and b={(A-B), we shall have

2
sin. A+sin. B=sin. 3(A+B) cos. :(A-B) (1)

2
sin. A-sin. B= sin. (A-B) cos. ](A+B) (2)

R

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

R

2
cos. B-cos. A sin. 3(A+B) sin. }(A-B) (4)

R (76.) Dividing formula (1) by (2), and considering that sin. a tang. a

(Art. 28), we have

R sin. A+sin. B sin. }(A+B) cos. 2(A-B)_tang. 1(A+B).

sin. A-sin. B*sin. (A-B) cos. 3(A+B) tang. }(A-B): that is,

The sum of the sines of two arcs is to their difference, as the tangent of half the sum of those arcs is to the tangent of half their difference.

cot. Dividing formula (3) by (4), and considering that

sin. R R

(Art. 28), we have tang cos. A+cos. B_ cos. }(A+B) cos. 3(A-B)_cot. 3(A+B)

cos. B-cos. Asin. I(A+B) sin. i(A,B)tang. }(A-B): that is

The sum of the cosines of two arcs is to their difference, as the cotangent of half the sum of those arcs is to the tangent

COS. a

COS.

From the first formula of Art. 74, by substituting A+B for

2, we have

we ob

> COS.

2 sin. }(A+B) Xcos. }(A+B) sin. (A+B)=

R Dividing formula (1), Art. 75, by this, we obtain sin. A+sin. B sin. }(A+B) cos. {(A,B)_cos. }(A-B) sin. (A+B)

sin. (A+B) cos. }(A+B) cos. 3(A+B)' that is,

The sum of the sines of two arcs is to the sine of their sum, as the cosine of half the difference of those arcs is to the cosine of half their sum.

If we divide equation (1), Art. 72, by equation (3), we shall have

sin. (a+b)_sin. a cos. b +cos. a sin. b

sin. (a-b)sin. a cos. b-cos. a sin. 6 By dividing both numerator and denominator of the second

tang. sin. member by cos. a cos. b, and substituting

for

R sin. (a+b) tang. a+tang. b tain sin. (ab)tang. a-tang. b; that is,

6 The sine of the sum of two arcs is to the sine of their difference, as the sum of the tangents of those arcs is to the difference of the tangents.

From equation (3), Art. 72, by dividing each member by cos a cos. b, we obtain sin. (ab)_sin. a cos. b-cos. a sin. b_tang. a-tang. 6

; b R cos, a cos. b

R? that is,

The sine of the difference of two arcs is to the product of their cosines, as the difference of their tangents is to the square of radius. (77.) Expressions for the tangents of arcs.

R sin. (a+b) If we take the expression tang. (a+b)=

(Art.

cos. (a+b) 28), and substitute for sin. (a+b) and cos. (a+b) their values given in Art. 72, we shall find

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

cos, a cos. b-sin. a sin. b

COS. A COS.

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tang. (a+b)=R-tang. a tang. b

cos. a tang. a

cos. b tang. b But sin a and sin. b=

(Art. 28). R

R If we substitute these values in the preceding equation, and divide all the terms by cos. a cos. b, we shall have

R' (tang. a+tang. b)

’ In like manner we shall find

R” (tang. a--tang. b) tang. (a−b)= R'+tang. a tang. b Suppose b=a, then

2R2 tang. a

R’-tang. a Suppose b=2a, then

R' (tang. a+tang. 2a)

R’-tang. a tang. 2a In the same manner we find

cot. a cot. b-R’ cot. (a+b)=

cot. b+cot. a

cot. a cot. b+R? cot. (ab)=

cot. b-cot. a (78.) When the three sides of a triangle are given, the angles may be found by the formula

tang. 2a=

tang. 3a=

sin. fA=RV

(SB)(S—c)

bc where S represents half the sum of the sides a, b, and c.

Demonstration.
Let ABC be any triangle; then (Geom., Prop. 12, B. IV.),
BO'=AB'+AC-2ABX AD.

C
AB'+AC-BC-
Hence, AD=

2AB
But in the right-angled triangle ACD,
we have

А D

B R: AC :: cos. A : AD.

RXAD Hence,

cos. A=

;

AC or, by substituting the value of AD,

AB'+AC-BC cos. A=RX

2ABXAC

Let a, b, c represent the sides opposite the angles A, B, C;

+c-a then,

cos. A=RX

2bc By Art. 74, we have 2 sin. 'JA=Ro-R cos. A. Substituting for cos. A its value given above, we obtain

b' +6 -- a 2bcta-b2 sin. 'JA=R'-R'X

=R'X 2bc

2bc Rox(a+b-c)(a+c-b)

2bc Put S={(a+b+c), and we obtain, after reduction,

(Sb) (S—c) sin. JA=RV

be In the same manner we find

S-a) (s—c) sin. IB=RV

}A

ac

(S-a) (S-6) sin. C=RV

ab Ex. 1. What are the angles of a plane triangle whose sides are 432, 543, and 654 ? Here S=814.5; S-b=382.5; S-c=271.5. log. 382.5

2.582631 log. 271.5

2.433770 log. b, 432

comp. 7.364516 log. c, 543

comp.

17.265200

2) 19.646117 sin. IA, 41° 42' 361".

9.823058. Angle A=83° 25' 13". In a similar manner we find the angle B=41° 0' 39", the angle C=55° 34' 8".

Ex. 2. What are the angles of a plane triangle whose sides are 245, 219, and 91 ?

(79.) On the computation of a table of sines, cosines, fc.

In computing a table of sines and cosines, we begin with finding the sine and cosine of one minute, and thence deduce the sines and cosines of larger arcs. The sine of so small an angle as one minute is nearly equal to the corresponding arc. The radius being taken as unity, the semicircumference is

and

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