TABLE For Reducing French Degrees, &c. to English. French Division: 400° in the circle, 100', in a degree, 100" in a minute. i CHAP. II. Expressions, for the Sines and Cosines of the Angles of a Triangle, in Terms of the Sides: for the Sine and Cosine of the Sum and Difference of two Arcs or Angles, &c. THE first step in this investigation will be made by the solution of the following Problem : PROBLEM 1. In an oblique-angled triangle, it is required to express the cosines of the angles in terms of the sides. Let the 3 angles be A, B, C, the opposite sides, a, b, c. Let the line between the vertex of the angle C and the point where a perpendicular from the vertex of A on the line a cuts a bep; then, pis called the cosine of C to the radius b, and (by p. 17.) = b.cos. C, when the radius = 1. Now, by Euclid, Book II, Prop. 12 and 13, If we investigate cos. B, and cos. A, the process will be exactly similar, and the result similar, that is From these expressions, the angles of a triangle may be found = a2 + b2 and c2 = a2 + b2 + 2 ap = a2 + b2 + 2ab.cos. (π - C) - 2ab.cos. C { since cos. (C) = - cos. C}; is the same, whether C be less or greater than a right angle. PROBLEM 2. Let it be required to express the sines of the angles in terms of the sides. cos. A By Euclid, Book I. Prop. 47. sin. A = (rad.)2 = (when rad3. = 1) 1 cos. A = (1 + cos. A) (1 - cos. A); since the difference of the squares of two quantities is equal to the product of their sum and difference; hence we may find the value of sin. A, by finding, from the preceding Problem, 1 + cos. A, and 1 - cos. A, and, then, by multiplying together those quantities. Hence multiplying, 1 + cos. A, and 1 - cos. A, sin. A = (a+b+c)(b+c-a)(a+b−c)(a+c-b); 462 2 or, since a+b-c=2(a+b+c_c), b+c-a=2(a+b+c-a), 22 {(a+b+c)(a+b+c)(a+b+c) 2 2 2 {(a+b+c)(a+b+c)(a+b+c)(a+b+c)}, 2 2 2 2 This is the expression for the sin.2 A, formed by means of that for cos. A. But, the expressions for cos. B, cos. C are precisely similar to that for cos. A, and, therefore, the sin.2 B and sin.2 C formed from them, by the same process, must be expressed by similar fractions; in which fractions, the numerators must, from the nature of their composition, be the same as the numerator for sin. A, and the denominators will be, respectively, a2c2, a2b. Let N2 represent the numerator, then 4bc: (b+c)2 a2 :: 2 (the diameter) : ver. sin. (180°-A) which is Halley's Theorem. Phil. Trans. No. 349. p. 466. Halley calls it "A new Theorem of good use in Trigonometry." or, if this equality be thrown into a proportion, sin. B: sin. C :: b : c. we have This relation, however, of the sines of the angles to the sides opposite, may be immediately deduced from (Fig. 3.); for the perpendicular (q) on a from the vertex of A, is the sine of B to the radius c, or q=c sin. B (radius = 1); similarly, q is the sine of C to the radius b, or, q=b sin. C (rad. 1.); ... since q=q, c sin. B = b sin. C. COR. 2. Hence the area of a triangle may be expressed in terms of its sides, for, (see Fig. 3.) area = a. perpac. sin. B = 2 2 PROBLEM 3. It is required to express the sine of the sum of two arcs, in terms of the sines and cosines of the simple arcs. From the two preceding Problems, Hence, sin. A. cos. B+cos. A. sin. B= N(a2+c2-b2+b2+c2-a2) N 2abc2 • But sin. C=, and sin. C=sin. { -(A+B) } =sin. (A+B), for A + B is supplement to - (A + B) as A + B + C = π. Hence sin. (A + B)=sin. A. cos. B+cos. A. sin. B. [1]* * It has been objected, and rightly, that this formula is not thus obtained |