where the upper signs refer to the angle XOP, and the lower signs to XOP'. 2. Given cot A = ing functions of A. The equation may be written cot A=-12. 5 We may then regard the point of reference as having its abscissa equal to 12 and its ordinate equal to 5, or as having its abscissa equal to -12 and its ordinate equal to 5; and there are consequently two angles, XOP and XOP', in the first and third quadrants respectively, either of which satisfies the given condition. Then OP=OP' = √0M2 + PM2 = √144 + 25 = 13. In each case find the values of the remaining functions: 57. We have from the general definitions of Art. 32, 58. To prove the formulæ, sin2 A+ cos2 A= 1, sec2 A=1+ tan2 A, csc2 A=1+ cot2 4, for any value of A. Since the distance of the point of reference is the hypotenuse of a right triangle whose sides are equal to the absolute values of the abscissa and ordinate, we have by Geometry, (ordinate)2+(abscissa)2 = (distance)2. Hence for any value of A, we have sin2 A + cos2 A=1, sec2 A=1+tan2 4. esc2 A= + cot2 A. IV. MISCELLANEOUS THEOREMS. TO EXPRESS EACH OF THE SIX PRINCIPAL FUNCTIONS IN TERMS OF THE OTHER FIVE. 60. The following table expresses the value of each of the six principal functions in terms of the other five: The reciprocal forms were proved in Art. 57; the others may be derived by aid of Arts. 57, 58, and 59, and are left as exercises for the student. As an illustration, we will give a proof of the formula Note. We suppose x to be expressed in circular measure (Art. 4). Let OPXP' be a circular sector; draw PT and P'T tangent to the arc at P and P', and join OT and ̧PP'. By Geometry, PT= PT. Then OT is perpendicular to PP' at its middle point M, and bisects the arc PP' at X. Let XOP= ≤ XOP' = x. By Geometry, arc PP' > chord PP', and < PTP'. or, by Art. 5, circ. meas. x > sinx, and < tan x. Representing the circular measure of x dividing through by sin x, we have (A) by x simply, and |