ક

Again, by the same supposition, Cor. 1 becomes

1: cos. A :: cos. B : ¦{cos. (A — B) + cos. (A + B)};

.. {cos. (A — B) + cos. (A + B)} = cos. A cos. B;

and similarly, Cor. 2 becomes

{cos. (AB) — cos. (A + B)} = sin. A sin. B;

adding and subtracting these, gives

cos. (AB) = cos. A cos. B + sin. A sin. B

(D),

In (E), let B = A, then cos. 2A=cos. A sin. A

COR. 5.-Using the same notation as in Cor. 4, the propósition becomes

1: sin. A :: cos. B: {sin. (A + B) + sin. (A — B)}; therefore {sin. (A + B) + sin. (A — B)} = sin. A cos. B ; multiplying by 2,

sin. (A + B) + sin. (A − B) = 2 sin. A cos. B ;

subtracting sin. (A – B) from both,

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

sin. (A

B);

If now in this formula A be put equal to 1', 2′, 3′, 4', &c., it becomes

2 sin. 1' cos. 1'-0 since sin. 0 = 0,

sin. 2'

In this manner a table of natural sines may be constructed extending through the whole quadrant, if the sine and cosine of one minute can be found. Now, in the Quadrature of the Circle, Proposition x. Cor. 1, it is shewn that Q2= (1 + P), where if P be the cosine of any arc, Q is the cosine of half the arc, therefore 1+ cos. A

A

; now it can easily be shewn that cos. 60°

, for Cor. 4 (C), sin. 2A = 2 sin. A cos. A; let A = 30°,

P

=

=

=

therefore sin. 60° 2 sin. 30 cos. 30, but cos. 30° sin. 60°, and sin. 30° =cos. 60°, hence sin. 60° =2 cos. 60° sin. 60°, and dividing by 2 sin. 60°, cos. 60° ; therefore cos. 30° can be found, and hence cos. 15° can be found. Proceeding in this manner, the cosine of an arc less than one minute may be found; and since (Trig. 10) sin.2A + cos.2A 1, sin. A = √1 — cos. A, from which the sine of the same arc may be found. Then since the sines of small arcs are very nearly proportional to their sines, the sine of one minute will be found to be: = 0002908882, when radius 1. The sine of 1' being found, the cos. 1'= √/1 — sin.3l′. In this way the cosine of 1' is found to be = 9999999577 = 1 0000000423. Now if this latter form of the cosine of 1' be substituted for it in formula (G) above, it becomes

COR. 6.-Using the same notation as in Cor. 4, it follows, from Cor. 3, that sin. (A + B) sin. (AB) = 2 cos. A sin. B in this formula, let A 60°, and remembering that cos. 60° =, and therefore 2 cos. 60° = = 1, it becomes

=

sin. (60° + B) — sin. (60° — B) = sin. B,

or sin. (60° + B) = sin. (60° - B) + sin. B,

which shews that when the sines have been calculated up to 60°, all those after may be found by taking the sine of an arc as much less than 60°, and adding to it the sine of the arc by which it exceeds 60°.

As shewn above, a table of the sines, and consequently of the cosines, of arcs of any number of degrees and minutes, from 0 to 90° may be constructed. And since tan. A =

=

sin. A

the

cos. A'

table of tangents is computed by dividing the sine of any arc by the cosine of the same arc. Again, since sec. A =

1

cos. A'

the

secants will be found by dividing 1 by the cosine of the arc whose secant is sought. Lastly, the versed sines are found by subtracting the cosines of the arcs from 1.

ANALYTICAL TRIGONOMETRY.

(1.) Again, resuming the results obtained in (Prop. xII.), and its first, second, and third corollaries, as collected in (Cor. 4); namely,

{sin. (A + B) + sin. (A — B)} = sin. A cos. B

(a),

If, in these four formulæ, A + B = S, and A B = D, then A = (SD), and B = (SD); substituting these values in the four formulæ above, and multiplying both sides by 2, they become

(2.) Also, since S and D are any two arcs, they may be represented by A and B; then the above four become

(3.) These four formulæ, expressed in words, prove the following important trigonometrical propositions:

(i.) The sum of the sines of two arcs is equal to twice the sine of half the sum of the arcs, multiplied into the cosine of half the difference of the arcs..

(k.) The difference of the sines of two arcs, is equal to twice the cosine of half the sum of the arcs, multiplied into the sine of half the difference of the arcs.

(1.) The sum of the cosines of two arcs is equal to twice the cosine of half the sum of the arcs, multiplied into the cosine of half the difference of the arcs.

(m.) The difference of the cosines of two arcs is equal to twice the sine of half the sum of the arcs, multiplied into the sine of half the difference of the arcs.

(4.) Again, resuming the expressions (A), (B), (D), and (E) in Cor. 4, Prop. XII.; namely,

(5.) In the expressions in (2), let B = 0, and remember that it has just been proved that sin. 0 = 0, and cos. 0 = 1; then each of the expressions (i) and (k) give (n) as above; while

(1) gives 1+ cos. A = 2 cos.2 A

...

(s),

(t).

(6.) Again, divide the expressions in (2) by one another, and they give

by dividing both numerator and denominator of the second side by 2 cos. (A + B) cos. (AB); but it was proved in (Prop. v. Trig.), that if A and B be two angles of a triangle,

and a and b the sides opposite to them, that a:b:: sin. A: sin. B, therefore (Book V. Prop. G)

which is the rule for finding the angles of a triangle when there are given two sides and the included angle; see Trig. Prop. VI.

In the last three expressions if B = 0, they become, remembering that by (p) and (r) sin. 0 = 0, and cos. 0 = 1,