In like manner, dividing (2) by (1), member by member, we have,

cot (90° + a)

= tan a.

Taking the reciprocals of both members of (2), we have (formulas 11 and 12, Art. 61),

sec (90° + a) = cosec a.

In like manner, taking the reciprocals of both members of (1), we have,

cosec (90° + a) = sec a.

Again, let M′′C = AM = a; then AM" 180° a. As before, the right-angled triangles triangles OP"M" and OPM may be proved equal in all respects, giving the numerical relations, P"M" = PM, and OP" = OP, and, by the application of the rules for signs, Art. 58, may be obtained, P"M" = PM, and OP" OP; hence,

= -

From these equations (1) and (2), and formulas (6), (7), (11), and (12), Art. 61, may be obtained, as before,

In like manner, the values of the several functions of the remaining arcs in question may be obtained in terms of functions of the arc α. Tabulating the results, we have the following

It will be observed that, when the arc is added to, or subtracted from, an even number of quadrants, the name of the function is the same in both columns; and when the arc is added to, or subtracted from, an odd number of quadrants, the names of the functions in the two columns are contrary: in all cases, the algebraic sign is determined by the rules already given (Art. 58).

By means of this table, we may find the functions of any arc in terms of the functions of an arc less than 90°. Thus,

PARTICULAR VALUES OF CERTAIN FUNCTIONS.

=

64. Let MAM' be any arc, denoted by ya, MM its chord, and OA a radius drawn perpendicular to M'M: then will PM M'M, and AM M'AM (B. III., P. VI.). But PM is the sine of AM, or, PM sin a: hence,

sin a = M'M;

M

that is, the sine of an arc is equal to one half the chord of twice the arc.

=

=

M'M will

Let M'AM 60°; then will AM 30°, and equal the radius, or 1 (B. V., P. IV.): hence, we have

sin 30°

that is, the sine of 30° is equal to half the radius.

Again, let M'AM

√2 (B. V., P. III.):

90°: then will AM 45°, and M'M = hence, we have

FORMULAS

EXPRESSING RELATIONS BETWEEN THE CIRCULAR FUNCTIONS OF DIFFERENT ARCS.

65. Let AB and BM represent two arcs, having the common radius 1; denote the first by

a, and the second by b; then, AM =
a + b.
From M draw PM perpendicular
to CA, and NM perpendicular to CB;
from N draw NP' perpendicular, and NL
parallel, to CA.

Then, by definition, we have

PM = sin (a + b), NM = sin b, and

From the figure, we have

PM = PL + LM.

N

B

P PA

CN cos b.

(1.)

From the right-angled triangle CP'N (Art. 37), we have

PL = cos b sin a = sin a cos b.

Since the triangle MLN is similar to CP′N (B. IV., P. XXI.), the angle LMN is equal to the angle P'CN; hence, from the right-angled triangle MLN, we have

LM NM cos a = sin b cos a = cos a sin b.

Substituting the values of PM, PL, and LM, in equation (1), we have

that is, the sine of the sum of two arcs is equal to the sine of the first into the cosine of the second, plus the cosine of the first into the sine of the second.

Since the above formula is true for any values of a and b, we may substitute -b for b; whence,

sin (a - b) sin a cos (-b) + cos a sin (— b);

that is, the sine of the difference of two arcs arcs is equal to the sine of the first into the cosine of the second, minus the cosine of the first into the sine of the second.

If, in formula (B), we substitute (90° have

sin (90°-a-b) = sin (90°—a) cos b-cos (90°-a) sin b; (2.)

hence, by substitution in equation (2), we have

that is, the cosine of the sum of two arcs is equal to the rectangle of their cosines, minus the rectangle of their sines.