By operating in the same way with the different expressions in Formula 3, we collect the following expressions :

It will furnish an easy and agreeable exercise to the student, to go over the different steps of the operations, by which these expressions are deduced.

If, in the second, third, and sixth of Formula 4, the arc B be taken = 0, we have

If the rectangle, or the product, of the first and second of Formula 1, be taken, we have sin A+B.sin A¬B=sin A. cos B+cos A, sin B. sin A. cos B cos A. sin B.

And as the rectangle, or the product, of the sum and difference of two quantities, is equal to the difference of their squares; sin A. cos B + cos A. sin B. sin A. cos B ços A. sin B, is equal

This expression converted into a proportion, gives us in words the following analogy:

As the sine of the sum of two arcs, is to the sum of their sines, so is the difference of their sines to the sine of their difference.

By an analogous process we deduce the following expression from the third and fourth of Formula 1.

Again, by dividing the first of Formula 1, by the third, and the second by the fourth, we have

Let both the numerator and denominator of the latter fraction, in each of these expressions, be divided by cos A. cos B, and the expressions become

In the first and third of Formula 1, and the first of Formula 6, let the arcs A and B be considered as equal, and the Formula will be reduced to the following, viz.

2 sin2 A, or 2 sin2 A = 1 − cos 2 A. And, if instead

A of 2 A we substitute A, we must substitute in the expression,

instead of A, and the expression will then become

1, or 2 cos2 A=1+ cos 2 A. And if, as above,

Recurring again to Formula 2, we have, by transposition,

(8)

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

B

(9)

In the first of those expressions, if A = 30°, 2 sin A will be equal to radius, or to unity. And in the last expression, if A = 60o, 2 cos A will be equal to radius.

Hence the expressions give

sin 30° + B = cos B sin 30°

B;

B.

(10)

and sin 60° + B = sin B + sin 60° By the former of these theorems, when the sines and cosines of all arcs below 30° have been computed, the sines may easily be continued to 60°; and then by the latter theorem, the sines of all arcs between 60°, and the quadrant will readily be obtained.

The foregoing formulae might all have been investigated geometrically, though they could not have been deduced with equal conciseness. We shall afterwards take occasion to shew how a few of them may be directly deduced from geometrical principles, but we shall leave the greater part of them to be so proved by the ingenious student whose taste or curiosity the exercise may gratify.

ON THE CONSTRUCTION OF TABLES OF NATURAL SINES, TANGENTS, &c.

A

be an arc whose cos is known, the cos of its half, may be

2

computed. Now 60° is an arc of that description; its cosine (Proposition 1. Trig.) being to radius unity. Hence cos 30°

cosine of 15°, 7° 30′, &c. may be computed; till, after twelve successive bisections, the cosine of 52" 44"" 3" 45' will be obtained. Then knowing the cos, we readily get the sine; for sin2 = rad 3— cos2,

rad+cos rad

cos; or sin =

1 + cos ·

1

cos, to radius unity. Thus therefore the sine of 52′′ 44′′′′ 31 45′ may be found.

Now as the sines of very small arcs vary nearly as the arcs themselves do, we have by analogy, as 52′′ 44′′ 3" 45' is to the sine of 52′′ 44′′ 31 45, so is l' to the sine of 1', and hence the sine of 1' may be computed.

Thus having the sine and cosine of 1', we obtain the sine of 2' from Formula 7.

sin 2 A

2 sin A. cos A,

or sin 2' 2 sin 1'. cos 1'.

For the sine of 3', and every succeeding minute, we have from Formula 9.

sin A + B = 2 sin A. Which expression if A be taken becomes

cos B

sin AB.

2′,3′, 4', &c. and B = 1′

Having computed the sines of a few minutes in this manner, the correctness of the computation may be ascertained by the application of another theorem, which flows immediately from the equation following Formula 5, viz.

sin AB sin A

sin B sin A + sin B: sin A + B.

Here let A= 3′ and B = 2′, and the expression is

sin 1': sin 3' sin 2 sin 3' + sin 2': sin 5'.

If the sine of 5', computed in this way, agree with the sine of

the same arc computed by the former, or any other independent method, the result may be considered as correct.

By the method here described the sines of all arcs may be computed, and their cosines may then be obtained by the formula, cos

Vi

1 + sin. 1 — sin, their tangents by tan

sin

COS

their cotan

We shall afterwards have occasion to investigate a formula by which the computation of the secants may be considerably facilitated.

PROPOSITION III.

In any right-angled plane triangle, as the hypothenuse is to either of the sides, so is the radius to the sine of the angle opposite to that side: and as either of the sides is to the other side, so is the radius to the tangent of the angle opposite to that side.

Let A B C be a plane triangle right angled at A; and from C as a centre with any radius C D, let the arc DE be described, meeting C B and C A in the points D and

E;
and from D and E, let D F and E G be drawn
perpendicular to A C, then D F will be the sine,
and E G the tangent of the arc E D, or of the
angle C.

B

DG

FE A

The triangles CF D, C E G, and C A B are similar, as they have the common angle C, and each of the CAB is a right angle. Hence C B

:

angles CFE, CEG, and BA:: CD: DF; or C B :

BA: rad sin C. And C A:AB::CE: E G, or C A:AB:: rad: tan C.

Cor. 1. C B . sin C = rad . B A,

and C A. tan C = rad . A B, or if rad be considered as unity B A C B. sin C, and B A = CA. tan C.

=

Hence C A. sec C = rad . C B ; or C B C A. sec C, radius being unity.

Cor. 4. To the hypothenuse, as a radius, each leg is the sine of its opposite angle; and to one of the legs as a radius, the other leg is the tangent of its opposite angle, and the hypothenuse is the secant of the same angle.