between the trigonometrical lines of two different arcs. They are introduced here because necessary for the solution of the few cases of plane triangles which remain. We shall first derive formulas by means of which, when the sines and cosines of two arcs are known, the sine and cosine of their sum or difference may be found. Thus if the sine and cosine of 30°, and also those of 20° be given, those of 50°: B 300 + 20° M E KP A Adding equations (1) and (2) and observing that EL+BL-BE -sin A B we have sin (a+b)=sin a cos b+ sin b cos a (3) 1, for a The second member must be understood to be divided by R= line cannot be equal to the sum of two surfaces (See Art. 64, IX.). Formula (3) is read thus: the sine of the sum of any two arcs is equal to the sine of the first into the cosine of the second plus the sine of the second into the cosine of the first, divided by radius. Again, CK COS CX CI= cos a cos b (4) (5) Subtracting (5) from (4), and observing that o K-E K = C E = COS A B (Art. 24), we have The second member of (6) must be understood to be divided by R or the first member to be multiplied by R to produce homogeneity. Formula (6) is read thus: the cosine of the sum of any two arcs is equal to the rectangle of their cosines minus the rectangle of their sines divided by radius. In formula (3) let a = 60° and b = 20° then by the first We shall derive expressions for sin (a—b) and cos (a —b) or the sine and cosine of the difference of two arcs in terms of the arcs themselves, by making, in the formulas just derived, for sin (a+b) and cos (a +b); b = b, observing that cos (—b) = cos b and sin ( — b) = =- - sin b (Art. 27). By this substitution there results * From each of the logs. 9.91052 and 9.25302, 10 must be rejected in order to pass from the table of logarithmic sines, &c., in which the radius is 10000000000, the logarithm of which is 10, to the table of natural sines, &c., in which the radius is 1. The characteristics of these logarithms would thus become I, but they may be read so as they stand according to Art. 54, Ex. 2. + The student may develope this as an exercise, and compare the result with cos 80° as given by the table of natural sines and cosines. Find the sine of 570 from sin of 15° 25882, and cos= 42066913, and cosine⚫74314. •96593, and sin Find the cosine of 9° from sine and cosine of 24° = •40674 and 91355 and sine and cosine 150 = •25882 and ⚫96593. These four formulas for the sine and cosine of the sum and difference of two arcs should be committed to memory, as they are constantly recurring in trigonometry, and in the higher analysis. The four may be expressed in two by the use of the double sign, thus in the higher analysis for expressing twice an arc in terms of the arc itself, by simply making b = a the result is the two terms of the second member becoming the same. We also get an analogous expression for the cosine of twice an arc by making b = a in formula (6) of the last article cos (a+b) expression is cos 2 a = cos1 α - sin' a = &c. (2) This Thus knowing the sine and cosine of 20°, these last two formulas would give us the sine and cosine of 40°. These two formulas may be modified so as to express the sine and cosine of an are in terms of half the arc, under which last form they are much used. This is accomplished by making aa, which is legitimate, since a is supposed to have no particular value; then 2 a becomes a and we have from (1) EXERCISES. Find the sine of 320 from the sine and cosine of 16°, the former being ⚫27564 and the latter ⚫96126. Find also the cosine of 320 from the same data. NOTE. This last is best done by observing that the product of the sum and difference is equal to the difference of the squares. 72. By means of this last, and a very simple formula depending upon the well-known property of the right angled triangle, that the square of the hypothenuse is equal to the sum of the squares of the other two sides, a formula expressing the value of the sine of half an arc in terms of the arc itself may be obtained. The formula depending upon the property of the right angled triangle, will be found by referring to the last diagram, in which the triangle C P M is right angled at P, whence (Geom. Th. 26). Introducing R into equation (4) according to the rule for homogeneity at Art. 31, and changing the order of the members it becomes Dividing by 2 and taking the square root of both members of this last equation, we have the formula required. 73. We resume the solution of triangles, having now a formula, by means of which we shall be able to derive an expression for one of the angles of a triangle in terms of the three sides; an expression which will be found much more convenient for the application of logarithms than that contained in Art. 69. putting в in the place of a in the formula for sina (formula (8) of the last article) we have substituting for cos в in this, its value in (1), we have reducing the terms under the radical to a common denominator, there results but the difference of the squares of two quantities is equal to the product of their sum and difference (Alg. Art. 13, Note 2), hence b2 — (a—c)' = (b + a −c) (b − a + c) substituting the second member of this in place of the first in the preceding equation, and separating the 4 of the denominator into two factors 2 × 2, we have representing b+a+c the sum of the three sides of the triangle by 8, the second members of the two last equalities become |