is called the sine of angle A, and is read, sine A; OP is called the cosine of angle A, and is read, cosine A; OP IV., MP III., is called the tangent of angle A, and is read, tangent A; OM OP MP is called the cosecant of angle A, and is read, cosecant A; is called the secant of angle A, and is read, secant A; OM VI. is called the cotangent of A, and is read, cotangent A. OM MP 29. The expressions sine A, cosine A, tangent A, cosecant A, secant A, cotangent A are abbreviated into sin A, cos A, tan A, csc A or cosec A, sec A, cot A, respectively. The powers of the trigonometric ratios are expressed as follows: and so on. (sin A)2 = sin A x sin A, is written sin' A; (cos A)3 = cos A⋅ cos A⋅ cos A, is written cos3 A; (tan A)" is written tan" A, The student must notice that "sin A" is a single symbol. It is the name of a number, or fraction, belonging to the angle A; and if it be at any time convenient, we may denote sin A by a single letter, such as s or x. Also sin2 A is an abbreviation for (sin A)2, that is, for (sin A) × (sin A). Such abbreviations are used because they are convenient. The student who succeeds in the study of trigonometry must commit the preceding definitions to memory. CHAPTER IV THE ACUTE ANGLE 30. Thus far we have placed no limitations on the magnitude of the angle under consideration at any time. In the present chapter we shall confine our attention to angles lying between 0° and 90°. We shall, in Chapter VII., return to the consideration of the general angle. 31. Given a right triangle OMP with the right angle at M. Then MOP is acute, as is also OPM. Let us consider angle MOP; call MOP, A; then, MP side of triangle opposite angle under consideration hypotenuse side of triangle adjacent to the angle under consideration. or, briefly, cos A = hypotenuse adjacent side. and similar expressions for the other ratios. EXERCISE. Write the trigonometrical ratios of angle P. 32. Assuming that the angle XOP is less than 90°, we shall show I. That so long as1 the angle remains unchanged, the ratios remain unchanged. II. That a small change in the angle produces a change in each of the ratios. I. Take any angle ROE; let P be any point in OE, one of the lines containing the angle, and let P', P" be any two points in OR, the other line containing the angle. Draw PM perpendicular to OR, and P'M', P"M" perpendiculars to OE. Then the three triangles OMP, OM'P', OM"P" each contain a right angle, and they have the angle at O common; therefore their third angles must be equal. Thus the three triangles are equiangular. Therefore the ratios MP M'P' M"P" are all equal. (Geom.) OP' OP OP But each of these ratios is opposite side with reference to the hypotenuse angle at 0; that is, they are each sin ROE. Thus, sin ROE is the same whatever be the position of the point P on either of the lines containing the angle ROE. Therefore sin ROE is always the same. II. Let XOP and XOP' be two angles nearly equal. (See Fig. 13). 1 We shall show (Art. 63) that this change must be, in general, less than 90°. However, our proof is rigorous for the proposition stated. EXERCISE I. Prove Proposition I. of this article for the remaining ratios. EXERCISE II. Prove Proposition II. of this article for the remaining ratios. 33. The student should observe carefully (i.) that each ratio, such as opposite side is a mere number; (ii.) that we have proved, in Art. 32, these ratios remain unchanged as long as the angle remains unchanged; (iii.) that if the angle be altered ever so slightly, there is a consequent alternation in the value of these ratios; (iv.) that since these ratios are all numbers, they are therefore algebraic quantities, and bence obey all the laws of ordinary algebra. (v.) that there is a right angle in the same triangle as the angle referred to. EXAMPLE. In Fig. 14, in which BDA is a right angle, find sin DBA and COS DBA. In this case (i.) DBA is the angle. (ii.) BDA is a right angle in the same triangle as the angle DBA. (iii.) DA is the side opposite DBA and is perpendicular to BD. (iv.) BA is the hypotenuse. (v.) BD is the adjacent side. 1 We shall use the sign #, <, >, to mean "is different from," "is not = 1. In the triangle ABC, C being a right angle, AB 25, CB = 16; find sin A, cos A. 2. If in the triangle ABC, C being a right angle, AC = 2, BC = 4, find sin B, cos B, and cot B. 3. Let ACB be any angle, and let ABC and BDC be right angles (see Fig. 14). Write down two values for each of the following ratios: (i.) sin ACB, (ii.) cos ACB, (iii.) tan ACB, (iv.) sin BAC, (v.) cos BAC, (vi.) tan BAC. 5. If ABC be any right-angled triangle with a right angle at C, and we let A, B, and C stand for the angles at A, B, and C respectively, and let a, b, and c be the measures of the sides opposite the angles A, B, and C respectively: C |