From the similar triangles ONM and OBT', we have,

From the right-angled triangle OAT, we have,

OT2

=

0д2 + AT2;

or, sec2a1+ tan2a. (13.)

From the right-angled triangle OBT'', we have,

OT2 = OB2 + BT2; or, co-sec2a = 1 + cot2a. (14.)

It is to be observed that Formulas (5), (7), (12), and (14), may be deduced from Formulas (4), (6), (11), and (13), by substituting 90° a, for and then making the proper reductions.

a,

ure, we shall discover the following relations, viz. :

FUNCTIONS OF ARCS

FORMED BY ADDING AN ARC TO, OR SUB

TRACTING IT FROM ANY NUMBER OF QUADRANTS.

63. Let α denote any are less than 90°. has preceded, we know that,

From what

By a simple inspection of the figure, observing the rule for signs, we deduce the following relations:

without reference to their signs: hence, we have, as before,

By a similar process, we may discuss the remaining arcs Collecting the results, we have the following

in question.

table:

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 2a, M'M its chord, and OA a radius drawn perpendicular to M'M: then will PM PM', = and AM AM'

(B. III., P. VI.). But PM is the sine of AM, or, PM = sin a; hence,

sin a = {M'M;

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

Let M'AM = 60°; then will AM = 30°, and M'M will equal the radius, or 1: hence, we have,

sin 30°;

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