From the similar triangles ONM and OBT', we have, ON : OM :: OB : OT", From the right-angled triangle OAT, we have, · (12.) OT2 OA2 + AT2; or, sec2α = 1 + tan2a. . (13.) From the right-angled triangle OBT", we have, 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 a, and then making the proper reductions. the arc AM" will be denoted bya (Art. 48). All the functions of AM"", will be the same as those of ABM""; that is, the functions of α are the same as the functions of 360° - a. From an inspection of the fig ure, we shall discover the following relations, viz.: M M' N A P' P N" MT"" M D FUNCTIONS OF ARCS FORMED BY ADDING AN ARC TO, OR SUBTRACTING IT FROM ANY NUMBER OF QUADRANTS. 63. Let a 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 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 in question. Collecting the results, we have the following 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 functions of an arc less than 90° 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° = 1 ; that is, the sine of 30° is equal to half the radius. Again, let M'AM = 90°: then will M'M = √√2 (B. V., P. III.): hence, we have, |