67. To trace the changes in the sign and magnitude of the tangent of an angle as the angle changes from 0° to 360°. Making the same construction as in Art. 65, As A changes from 90° to 180°, QP revolves from QN to QW; .. PM changes from QN to 0 and is positive, As A changes from 180° to 270°, QP revolves from QW to QS; .. PM changes from 0 to QS and is negative, As A changes from 270° to 360°, QP revolves from QS to QE; .. PM changes from QS to 0 and is negative, 68. The changes of the cosecant, secant and cotangent should be traced for himself by the student for practice. 69. We now present the changes of the Trigonometrical Ratios in a convenient tabular form. Columns 1, 3, 5, 7, 9 give the values of the ratios for the particular values of the angle placed above the columns. Columns 2, 4, 6, 8 give the signs of the ratios as the angle passes from 0 to 90°, from 90° to 180°, from 180° to 270°, and from 270° to 360°. 70. From the results given in this table we are led to the following conclusion, which will be more fully explained hereafter. If the value of a Trigonometrical Ratio be given, we cannot fix on one angle to which it exclusively belongs. Thus if the given value of sin A 1 be we know, since sin A 2' passes through all values from 0 to 1 as A increases from 0° to 90o, that one value of A lies between 0° and 90o. But since we also know that the value of sin A passes through all values between 1 and 0 as A increases from 90° to 180°, it is evident that there is another value of A between 90° and 180° for which CHAPTER VI. 71. ON RATIOS OF ANGLES IN THE FIRST QUADRANT. We have now to treat of the values of the Trigonometrical Ratios for some particular angles in the first quadrant. These angles, which we shall take in their proper order, as they are traced out by the revolving line, are 0o, 30°, 45°, 60°, 90°. 72. The signs of all the ratios for an angle in the first quadrant are positive. 73. To find the Trigonometrical Ratios for an angle of 0o. We have already proved in the preceding Chapter that sin 0° = 0, cosec (0)o = ∞, cos 0o = 1, tan 0o = 0. 74. To find the Trigonometrical Ratios for an angle of 30o. Let QP revolving from the position QE describe an angle EQP equal to one-third of a right angle, that is an angle of 30o. Draw the chord PMP at right angles to QE, and join QP'. Then angle QPP = QPP' = 90° - PQM = 60°. Thus PQP' is an equilateral triangle, and QM bisects PP'; ... QP=2PM. Let the measure of PM be m. Then the measure of QP is 2m. And the measure of QM is √(4m2 — m3) = √(3m3) = m. √3. 75. To find the Trigonometrical Ratios for an angle of 45o. Let QP revolving from the position QE describe an angle EQP equal to half a right angle, that is an angle of 45o. Draw PM at right angles to QE. Then since PQM and QPM are together equal to a right angle, and PQM is half a right angle, QPM is also half a right angle. Thus PQM is an isosceles triangle, and QM = MP. Let the measure of QM be m. Then the measure of PM is m. And the measure of QP is √(m3 + m3) = √(2m3) = m √2. So also cosec 45° = √2, sec 45°/2, cot 45° = 1. 76. To find the Trigonometrical Ratios for an angle of 60o. Let QP revolving from the position QE describe an angle EQP equal to two-thirds of a right angle, that is an angle of 60o. Draw PM at right angles to QE and join PE. Then PQE is an equilateral triangle and PM bisects QE; Let the measure of QM be m. Then the measure of QP is 2m. And the measure of PM is √(4m2 — m3) = √(3m3) = m √3. |