ceding equations, that one should be taken which applies to the greater angle, whether that be the particular angle required or not; because the equations bring out the cosines to the angles; and the cosines to very small arcs vary so slowly, that it may be impossible to decide, with sufficient numerical accuracy, to what particular arc the cosine belongs. For instance, the cosine 9.999999, carried to the table, applies to several arcs; and, of course, we should not know which one to take; but this difficulty does not exist when the angle is large; therefore, compute the largest angle first, and then compute the other angles by Proposition 4.

But we can deduce an expression for the sine of any of the angles, as well as the cosine. It is done as follows:

EQUATIONS FOR THE SINES OF THE ANGLES. Resuming equation (m), and considering radius unity, we have

Equating the second members of (1) and (2),

By taking equation (p), and proceeding in the same manner, we have

The preceding results are for radius unity; for any other radius, we must multiply by the number of units in such radius. For the radius of the tables we write R; and if we put it under the radical sign, we must write R2; hence, for the sines corresponding with our logarithmic table, we must write the equations thus,

A large angle should not be determined by these equations, for the same reason that a small angle should not be determined from an equation expressing the cosine.

In practice, the equations for cosine are more generally used, because more easily applied.

The formule which we have thus analytically developed, express nearly all the important relations between the sines, cosines, and tangents of arcs or angles; and we have also demonstrated all the theorems required for the determination of the unknown parts of any plane triangle, three of the parts of which are given, one at least being a side.

Such relations might be indefinitely multiplied, but those already established are sufficient for most practical purposes, and when others are required, no difficulty will be found in deducing them from these.

The following geometrical demonstrations of many of the preceding relations, are offered, in the belief that they will prove useful disciplinary exercises to the student.

1st. Let the arc AD=A; then DG-sin.A; CG=cos.A; DI=sin.†A; AD=2sin.†A; CI=cos.†A; CI DO; and DB=2D0=2cos.A.

The angle, DBA, is measured by

one half the arc AD; that is, by †A.

Also,

ADG=DBA=†A.

Now, in the triangle, BDG, we have sin.DBG: DG=sin.90° : BD. That is, sin.A: sin.A 1: 2cos.A.

Or,

sin.A=2sin.A cos.†A;

which corresponds to equation (30).

In the same triangle,

sin.90° BD=sin.BDG: BG; and sin.BDG=cos.DBG.

That is, 1: 2cos.A=cos.A: 1+cos.A.

Or,

2cos.4-1+cos.A, same as equation (34).

In the triangle, DGA, we have,

sin.90°: AD= sin. GDA: GA.

That is, 1 : 2sin.†A=sin.†A : 1—cos.A.

Or,

2sin.' A1-cos. A, same as equation (35)

By similar triangles, we have,

That is,
Or,

BA: AD=AD: AG.

2: 2sin.42sin.A versed sin.A. versed sin.A = 2sin.' A.

Fm-mD=CH- cos.B; mn= cos.A;

cos.A+ cos.B Fn;

=

E/H

therefore, Fm+mn

cos.B cos.A= =nD;

But the chord, FB, is twice the sine of arc FB;

by one half of the arc BD; that is, by

two triangles, GnD and FnB, are similar.

A-B
2

and the

The angle, GFn, is measured by (4±3).

A+ B
2

In the triangle, FBG, Fn is drawn from an angle per

pendicular to the opposite side; therefore, by Proposition 5, we have,

Gn: nB tan. GFn: tan.BFn.

=

That is, sin.A+sin.B: sin.A-sin. B-tan.
1. (A + B) :

2

: tan.

A+B

In the triangle, GnD, we have,

sin.90°: DG = sin.nDG: Gn; sin.nDGcos.nGD.

That is, 1 : 2sin. (4+

Or,

sin. A+ sin. B=2sin.

(A+B) COB. (4B),

A

the same as equation (15).

8d. In the triangle, FnB, we have,

sin.90 FB sin. BFn: Bn.

the same as equation (16).

4th. In the triangle, FBn, we have,

sin.90: FB cos.BFn: Fn.

That is, 1: 2cos.

A+B

(4±3) = cos.(4–3) :

2

: cos.A+cos.B.

A

Or, cos.A + cos.B = 2cos. (A+B) cos. (4B), the

same as equation (17).

2

5th. In the triangle, GnD, we have,

sin.90° GD = sin.nGD: nD.

2

That is, 1 : 2sin. (A+B).

A-B

= sin.

: cos.B―cos.A,

2

the same as equation (18).

6th. In the triangle, FGn, we have,

sin. GFn: Gn cos. GFn: Fn.