Let OP, OP, OP, OP, represent the position of the revolving line at any period of revolution in the several quadrants respectively, And let PN, PN, PN, PN be the respective perpendiculars from the end of the revolving line upon the initial line. Then P,N, P,N, PN, PN are respectively the perpendiculars corresponding to the angles generated. Also, ON,, ON, ON, ON, are respectively the bases of the right-angled triangles with respect to the angles in question. We have then in the second quadrant— Sin AOP, = PN, cos AOP, OP = ON, It is therefore evident that the relations between the trigonometrical ratios, which were proved to exist in Art. 7, also hold for angles in the second quadrant—that is, angles between 90° and 180°. And in the same way we may show that they hold for angles in the third, fourth, or any quadrant. And again, if we suppose the line to revolve in a negative direction, and take the position OP', we shall have P'N' the perpendicular corresponding to the negative angle AOP', and ON' the base. And the relations proved in Art. 7 may be also similarly proved to exist here. Hence the relations proved in Art. 7 hold for any angles whatever. Changes of Magnitude and Sign of the Trigonometrical Ratios. 11. Let OP, OP, OP, OP, be positions of the revolving line in the several quadrants respectively; PN1, P ̧NË PN, PN, the respective perpendiculars; and ON1, ON2, ON, ON, the bases of the corresponding right-angled triangles. At the commencement of the motion of the revolving line, the angle AOP1 = 0°; Also, the perpendicular P1N1 = 0, And the base ON = Hence, we have OP1 1 tan 0° = 0. As the revolving line moves from OA towards OB, P1N1 increases and ON, diminishes; and when it arrives at ÕB, we have P1N1 OP1,and ON1 = 0. But the angle generated = is now a right angle. Hence we have— Hence, as the angle increases from 0° to 90°— The sine changes in magnitude from 0 to 1 and is +. The cosine changes in magnitude from 1 to 0 and is +. The tangent changes in magnitude from 0 to ∞ and is +. (2.) In the second quadrant Here the perpendicular P2N, is +, and the base ON, is 2 *The student ought properly to look upon the values 0, 1, 0 here obtained as the limiting values of the sine, cosine, and tangent respectively, when the angle is indefinitely diminished. Hence the sine during the second quadrant is +, the cosine is and the tangent is 2 Again, as the revolving line moves from OB to OA', the perpendicular P,N, diminishes until it becomes zero. Also, the base ON, increases in magnitude, until it finally coincides with OA', and .. - OP2 But the angle now described is 180°. = Hence in the second quadrant— = The sine changes in magnitude from 1 to 0, and is positive. The cosine changes in magnitude from 0 to 1, and is negative. The tangent changes in magnitude from ∞ to 0, and is negative. And in the same way may we trace the changes of magnitude and sign in the third and fourth quadrants. Thus we shall find (3.) In the third quadrant The sine changes in magnitude from 0 to 1, and is negative. The cosine changes in magnitude from 1 to 0, and is negative. The tangent changes in magnitude from 0 to ∞o, and is positive. (4.) In the fourth quadrant— The sine changes in magnitude from 1 to 0, and is negative. The cosine changes in magnitude from 0 to 1, and is positive. The tangent changes in magnitude from co to 0, and is negative. Moreover, as the cosecant, secant, and cotangent are respectively the reciprocals of the sine, cosine, and tangent, it follows that their signs will follow respectively the latter, and that their magnitudes will be their reciprocals. CHAPTER IV. TRIGONOMETRICAL RATIOS CONTINUED. ARITHMETICAL VALUES OF THE TRIGONOMETRICAL RATIOS OF 30°, 45°, 60°, &c. Using the same figure as in Art. 5, we have— = 45°, then also ZAPM, ZAPM 90° - 45° 45°. And hence PAM = = 14. Ratios of 30° and 60°. In the same diagram, suppose PAM Hence, if we conceive another triangle equal in every respect to APM to be described on the other side of AM, the whole would form an equilateral triangle whose side is AP. Hence, PM = AP. |