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

From the right-angled triangle OAT, we have,

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

=

OT12 OB2 + BT2; or, co-sec2a = 1 + cot2a.

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 a, and then making the proper reductions.

Collecting the preceding Formulas, we have the following table :

From an inspection of the fig

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

A

FUNCTIONS OF ARCS FORMED BY ADDING AN ARC TO, OR SUBTRACTING IT FROM ANY NUMBER OF QUADRANTS.

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

From what

Now, suppose that BM' = a, then will AM' = 90° + a.

We see from the figure that,

By a simple inspection of the figure, observing the rul 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 are 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.

61. 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;

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.

Again, let M'AM = 90° : then will AM
M'M = √2 (B. V., P. III.): hence, we have,

=