Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

sin 2a:

sin (a-b)

cos (a-b)=

R

These are the formulas which it was required to find.

The preceding demonstration may seem defective in pomt of generality, since, in the figure which we have followed, the ares a and b, and even a+b, are supposed to be less than 90 But first the demonstration is easily extended to the case in which a and b being less than 90°, their sum a+b is greater than 90°. Then the point F would fall on the prolongation of AC, and the only change required in the demonstration would be that of taking cos (a+b)=-CF'; but as we should, at the same time, have CF'I'L-CK', it would still follow that cos (a+b)=CK'—I'L', or R cos (a+b)=cos a cos b—sin a sin b. And whatever be the values of the arcs a and b, it is easily shown that the formulas are true: hence we may regard them as established for all arcs. We will repeat and number the formulas for the purpose of more convenient reference. sin a cos b+sin b cos a

sin (a+b)

(1.).

sin (a-b)=

=

sin a

whence

=

2 sin a cos a
R

sin a cos b—sin b, cos a

R

cos a cos b+sin a sin b

[ocr errors]

cos (a+b)=

cos (a-b):

XX. If, in the formulas of the preceding Article, we make d. =a, the first and the third will give

cos2 a-sin2 a

ק

R

sin a cos b-sin b cos a

R
cos a cos b-sin a sin b

R
cos a cos b+ sin a sin b
R

Los 2a= 9

(2.).

(3.)

[ocr errors]

(4.)

Ꭱ formulas which evable is to find the sine and cosine of the double arc, when we know the sine and cosine of the arc itself. To express the sin " and cos a in terms of a, put ja for a, and we have

cos2

2 cos2 a―
R

a-sin2 a

2 sin la cos a
R

R

To find the sine and cosine of a in terms of a, take th equations

-R

cos2 fa+sin2 a=R2, and cos2a-sin2 a R cos a, there results by adding and subtracting

cos2 a R2 + R cos a, and sin2 ja=R2-R cos a:

sin a= √(}R2—R cos a)=√2R2—2R cos a.
cos ļa=✓ (R2 + }R cos a)=√2R2 4 zR cos o

If we put 2a in the place of a, we shall have,

sin a=√(RR cos 2a)=√2R2-2R cos 2a. cos a= √(}R2+ R cos 2a) = √2R2+2R cos 2u. Making, in the two last formulas, a=45°, gives cos 2a=0, and sin 45°= √}R2=R✓; and also, cos 45°√R2=R√ }. Next, make a=22° 30′, which gives cos 2a=R√, and we have sin 22° 30' R√√) and cos 22° 30′ =R√(+}√}).

XXI. If we multiply together formulas (1.) and (2.) Art. XIX. and substitute for cos2 a, R2-sin2 a, and for cos2 b, R2-sin2 b; we shall obtain, after reducing and dividing by R2, sin (a+b) sin (ab)=sin2 a-sin2b= (sina+sin b) (sin a—sin b). or, sin (ab) sin a-sin b :: sin a+sin b: sin (a+b).

XXII. The formulas of Art. XIX. 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—b)=

sin (a+b) sin (a—b).

cos (a+b)+cos (a—b):

cos (a—b)—cos (a+b)=sin a sin b.

and which serve to change a product of several sines or cosires into linear sines or cosines, that is, into sines and cosines multiplied only by constant quantities.

XXIII. If in these formulas we put a+b=p, a—b=q, which gives a=

p+q

b

we shall find

2

[ocr errors]

2

sin p+sin q=

cosp+cos q=

[ocr errors]

cos q-cosp=

2

R

2

R

2

R

2

sin a cos b.

sin b cos a.

-cos a cos b.

2

sin p-sin q=sin (p—q) cos § (p+q) (2.)

2

sin (p+q) cos (p−q) (1.)

R

2

1

cos (p+q) cos } (p—q) (3.)

R

2

sin † (p+q) sin † (p—q) (4.)

R

[blocks in formation]

cos-p=

sin P

R+cos p
sin P
R-cos p

tang p
R

cot P

R

: hence

R

cot P R

tangp:

=

}

==

formulas which are often employed in trigonometrical calculations for reducing two terms to a single one.

R

XXIV. From the first four formulas of Art XXIII. and the first sin a tang a COS a R

R

cot a

of Art. XX., dividing, and considering that
we derive the following:

sin p+si q_sin } (p+q) cos † (p—q) ___ tang } (p+q)
sin p-sin q cos (p+q) sin (p−q) tang } (p—q)
sin p+sin q_sin } (p+q)
sin (p+q)_tang (p+q)
cos p+cos q cos &p+q)
sin p+sin q_cos (P-q) cot (p-q)
cos q-cos p sin (p—q)
sin p-sin q_sin } (p—q) __ tang} (p-
cos p+cos q cos (p-q)
R
sin p-sin q cos (p+q)_cot (p+q)
}
cos q-cos p sin (p+q) R

R

cos p+cos q_cos † (p+q) cos § (p—q) cot (p+q)
}
cos q-cos p sin (p+q) sin (p-q) tang (p—q)
≥ § (p—q)
sin p+sin q 2sin(p+q) cos (p-q) cos (p-q)
sin (p + q) ̄ ̄ 2sin (p+q) cos 1⁄2 (p+q) cos (p+q)

1⁄2

=

sin p—sin q_2sin } (p—q) cos } (p+q) _ sin ‡ (p—q) (p+q) 2sin(p+q) cos (p+q) sin (p+q) 1⁄2

sin

Formulas which are the expression of so many theorems From the first, it follows that the sum of the sines of two arcs i to the difference of these sines, as the tangent of half the sum of the arcs is to the tangent of half their difference

XXV. In order likewise to develop some formulas relative to tangents, let us consider the expression R sin (a+b) in which by substituting the values cos (a+b)

tang (a+b)=.

9

of sin (a+b) and cos (a+b), we shall find
R (sin a cos b+sin b cos a)

tang (a+b)=

=

cos a cos b―sin b sin a cos a tang a

and sin b

9

R

R

Now we have sin a= substitute these values, dividing all the terms by cos a cos b; we shall have

tang (a+b) =

R2 (tang a+tang b)
R2-tang a tang b

;

which is the value of the tangent of the sum of two arcs, expressed by the tangents of each of these arcs. For the tangent of their difference, we should in like manner find

R2 (tang a―tang b) tang (a—b)=R2+tang a tang b.

Suppose b=a; for the duplication of the arcs, we shall have the formula

tang 2a=

Suppose b=2a; for their

mula

tang 3 a=

cos b tang b

in which, substituting the tang 3 a=

2 R2 tang a

R2-tanga triplication, we shall have the for

;

R2 (tang a+tang 2 a)
R2-tang a tang 2 a
value of tang 2 a, we shall have
3R2 tang a-tang 3a

R2-3 tang 2a.

:

tang 2 a=

XXVI. Scholium. The radius R being entirely arbitrary, is generally taken equal to 1, in which case it does not appear in the trigonometrical formulas. For example the expression for the tangent of twice an arc when R=1, becomes,

2 tang a
1-tang2 a.

If we have an analytical formula calculated to the radius of 1, and wish to apply it to another circle in which the radius is R, we must multiply each term by such a power of R as will make all the terms homogeneous: that is, so that each shall contain the same number of literal factors.

CONSTRUCTION AND DESCRIPTION OF THE TABLES.

XXVII. If the radius of a circle is taken equal to 1, and the lengths of the lines representing the sines, cosines, tangents, cotangents, &c. for every minute of the quadrant be calculated, and written in a table, this would be a table of natural sines, cosines, &c.

XXVIII. If such a table were known, it would be easy to calculate a table of sines, &c. to any other radius; since, in different circles, the sines, cosines, &c. of arcs containing the same number of degrees, are to each other as their radii.

XXIX. If the trigonometrical lines themselves were used, it would be necessary, in the calculations, to perform the operations of multiplication and division. To avoid so tedious a method of calculation, we use the logarithms of the sines, cosines, &c.; so that the tables in common use show the values of the logarithms of the sines, cosines, tangents, cotangents, &c. for each degree and minute of the quadrant, calculated to a given radius. This radius is 10,000,000,000, and consequently its logarithm is 10.

XXX. Let us glance for a moment at one of the methods of calculating a table of natural sines.

The radius of a circle being 1, the semi-circumference is known to be 3.14159265358979. This being divided successively, by 180 and 60, or at once by 10800, gives .0002908882086657, for the arc of 1 minute. Of so small an arc the sine, chord, and arc, differ almost imperceptibly from the ratio of equality; so that the first ten of the preceding figures, that is, .0002908882 may be regarded as the sine of 1'; and in fact the sine given in the tables which run to seven places of figures is .0002909. By Art. XVI. we have for any arc, cos= √(1--sin3). This theorem gives, in the present case, cos l'=.9999999577. Then by Art. XXII. we shall have

2 cos l'x sin l'-sin 0'sin 2'=.0005817764
2 cos l'x sin 2'-sin 1'-sin 3'=.0008726646
2 cos l'x sin 3'-sin 2'sin 4'-.0011635526
2 cos l'x sin 4'-sin 3'-sin 5'-.0014544407
2 cos l'x sin 5'-sin 4'-sin 6'=.0017453284
&c.

&c.

&c.

Thus may the work be continued to any extent, the whole difficulty consisting in the multiplication of each successive result by the quantity 2 cos 1'1.9999999154.

« ΠροηγούμενηΣυνέχεια »