HOMOGENEITY OF TERMS. = = cot tang cos cos sin sin (x 1) (y 1) (≈ 1) (p + q) (p-q) These formulas are the algebraic enunciations of so many theorems. The first expresses that, the sum of the sines of two arcs is to the difference of those sines, as the tangent of half the sum of the arcs is to the tangent of half their difference. (p − q) ― (p + q) (p − q) - 4) (p + q) 87. An expression is said to be homogeneous, when each of its terms contains the same number of literal factors. Thus, sin2 a + cos2 a = R2 (1) is homogeneous, since each term contains two literal factors. If we suppose R = 1, we have, sin a + cos a = 1. . (2) This equation merely expresses the numerical relation between the values of sin a, cos a, and unity. If we pass from the radius 1 to any other radius, as R, it becomes necessary to replace these abstract numbers by their corresponding literal factors. For this, we must observe, that the radius of a circle bears the same ratio to any one of the functions of an arc, (the sine for example,) as the radius of any other circle, to the corresponding function of a similar arc in that circle. For example, 1 : sin a :: R: ·9 1 sin a sin a R R hence, in which the sin a, in the first member, is calculated to the radius 1, and in the second, to the radius R. If, now, we substitute this value of sin a to radius 1, in equation (2), we have, + = sin a; sin a + cos2 a = COS a R2, R 1 ; or, an expression which is homogeneous: and any expression may be made homogeneous in the same manner; or, it may be made so, by simply multiplying each term by such a power of R as shall give the same number of linear factors in all the terms. 88. Since the sine of an are divided by the radius is equal to the sine of another are containing an equal number of degrees divided by its radius, we may, if we please, define the sine of an arc to be the ratio of the radius to the perpendicular let fall from one extremity of the are on a diameter passing through the other extremity. Giving similar definitions to the other functions of the arc, each will have a corresponding function in either angle of a triangle. For, if in a right angled triangle, we let = A right angle; B = angle at base; C = vertical angle; α= hypothenuse; c = base; b = perpendicular, we may deduce all the functions of the angle without any reference to the circle. For, let us call, by definition, b -9 a sin B = tan B = sec B = sin A a sin A b с = α α с cos B = cot B= cosec B = Each of these expressions, regarded as a ratio, is a mere abstract number. If we make the hypothenuse a = 1, the abstract numbers will then represent parts of a rightangled triangle, or the corresponding functions of a circle whose radius is unity. Formulas relating to Triangles. 89. Let ACB be any triangle, and designate the sides by the letters a, b, c; then (Art. 21), sin B b ; C sin C с с α b = (1) A that is, the sines of the angles are to each other as their opposite sides. 90. We also have (Art. 22), a+b : α · b :: tan (A+B) : tan (A - B): that is, the sum of any two sides is to their difference, as the tangent of half the sum of the opposite angles to the tangent of half their difference. 91. In case of a right-angled triangle, in which the right angle is B, we have (Art. 24), 1 : tan A :: C : a; hence, a = c tan A, . (2) And again (Art. 25), 1 : cos A :: c; hence, cb cos A, . (3) 92. There is but one additional case, that in which the three sides are given to find an angle. Let ACB be any triangle, and CD a perpendicular upon the base. Then, whether the perpendicular falls without or within the triangle, we shall have (B. IV., P. 12), CBAC2 + AB- 2AB X AD. hence, 2sin2 4 cos A = But, and representing the sides by letters, and substituting for AD, its value, we have, sin A 1 + cos A If we now substitute for cos A, its value from formula (Art. 81), we shall have, = = b2 + c2 — a2 - 2bc 2bc (b2 + c2 - a2), 2bc a2 - b2 - c2+2bc 2bc b2 + c2 — a2 " 2bc = a2 − (b − c)2, 2bc (a+bc) (a+c− b), + b sin 4 = =V = 2bc - 93. If we add 1 to each above, in which we have the have, b If now, 1⁄2 (a + b + c) s, we have a + b + c = c) (a + c − b). 4bc с BD b) (s — c), bc a 2bc+b+c2-a2 2bc C = a B member of the equation value of cos A, we shall (b + c)2 - a2 (b + c + a) (b + c − a); and, = 2bc 1 + cos A cos | A = Substituting for 1+ cos A, its value (Art. 82), and reduc ing, we have, = tan A = √ tan B = V 2s (s-a). bc 94. If, now, we recollect that the tangent is equal to the sine divided by the cosine (Art. 47), we have, tan C = √s (s — a) bc (s — b) (s — c). s (s and observing that the same formula applies equally to either of the other angles we have, a) (sc), s (s - b) a) (s - b) s (s - CONSTRUCTION OF TRIGONOMETRICAL TABLES. 95. If the radius of a circle is taken equal to 1, and the lengths of the lines representing the sines, cosines, tangents, cotangents, &c., for every minute of the quadrant be calculated, and written in a table, this would be a table of natural sines, cosines, &c. 96. If such a table were known, it would be easy to calculate a table of sines, &c., to any other radius; since, in different circles, the sines, cosines, &c., of arcs containing the same number of degrees, are to each other as their radii (Art. 87). 97. Let us glance for a moment at some of the methods of calculating a table of natural sines. When the radius of a circle is 1, the semi-circumfer |