220. b. By comparing the expressions for the sines, and cosines, with those for the tangents and cotangents, a great variety of formulæ may be obtained. Thus, the tangent of the sum or the difference of two arcs, may be expressed in terms of the cotangent. Putting radius =1, we have (Arts. 93, 220.) sin (a+b)___sin a cos b+sin b cos a Dividing the last member of the equation, in the first place by cos a cos b, as in Art. 217, and then by sin a sin b, we have sin (a+b) tan a+tan b cot b+cot a = sin (a--b) tan a-tan bcot b-cot a In a similar manner, dividing the expressions for the cosines, in the first place by sin b cos a, and then by sin a cos b, we obtain cos (a+b)___cot b-tan a cot a-tan b == cos (a--b) cot b+tan a cot a+tan b Dividing the numerator and denominator of the expression for the tangent of a, (Art. 218.) by tan ja, we have These formulæ may be multiplied almost indefinitely, by combining the expressions for the sines, tangents, &c. The following are put down without demonstrations, for the exer Expression for the area of a triangle, in terms of the sides. 221. Let the sides of the triangle ABC (Fig. 23.) be expressed by a, b, and c, the perpendicular CD) by p, the segment AD by d, and the area by S. Then a2bc2-2cd, (Euc. 13. 2.) Reducing the fraction, (Alg. 150.) and extracting the root of both sides, This gives the length of the perpendicular, in terms of the sides of the triangle. But the area is equal to the product of the base into half the perpendicular height. (Alg. 518.) that is, Step=v4bc—(b2+c2—a2)2 Here we have an expression for the area, in terms of the sides. But this may be reduced to a form much better adapted to arithmetical computation. It will be seen, that the quantities 4b'c', and (b+c-a) are both squares; and that the whole expression under the radical sign is the difference of these squares. But the difference of two squares is equal to the product of the sum and difference of their roots. (Alg. 235.) Therefore, 4b'c'—(b2+c2--a2)2 may be resolved into the two factors, 2bc+(b2+c2-a) which is equal to (b+c)2-a2 Each of these also, as will be seen in the expressions on the right, is the difference of two squares; and may, on the same principle, be resolved into factors, so that, (b+c)2-a2=(b+c+a)×(b+c-a) {" Substituting, then, these four factors, in the place of the quantity which has been resolved into them, we have, S=÷v(b+c+a)×(b+c−a)×(a+b—c)×(a−b+c The expression for the perpendicular is the same, when one of the angles is obtuse, as in Fig. 24. Let AD = d. Then a2=b2+c2+2cd. (Euc. 12, 2.) And d= -b2 —c2+a2 2c 4c2 -(Alg.219.) Therefore, d2. 4c2 And p v4b2c2—(b2+c2—a2)2 as above. 2c Here it will be observed, that all the three sides, a, b, and c, are in each of these factors. Let h=(a+b+c) half the sum of the sides. Then S=vhx(h−a)x(h−b)x(h−c) 222. For finding the area of a triangle, then, when the three sides are given, we have this general rule; From half the sum of the sides, subtract each side severally; multiply together the half sum and the three remainders; and extract the square root of the product. SECTION VIII. COMPUTATION OF THE CANON. ART. 223. THE trigonometrical canon is a set of tables containing the sines, cosines, tangents, &c., to every degree and minute of the quadrant. In the computation of these tables, it is common to find, in the first place, the sine and cosine of one minute; and then, by successive additions and multiplications, the sines, cosines, &c., of the larger arcs. For this purpose, it will be proper to begin with an arc, whose sign or cosine is a known portion of the radius. The cosine of 60° is equal to half radius. (Art. 96. Cor.) A formula has been given, (Art. 210,) by which, when the cosine of an arc is known, the cosine of half that arc may be obtained. 30° 15° By successive bisections of 60°, we have the arcs 0° 28' 711 30!!! 0 14 3 45 If the radius be 1, and if a=60°, b-30°, c-15°, &c.; then |