The following formulae, the first three for right angled triangles, will often be of use, and may be easily demonstrated by the student. Let a and b be any two quantities, of which a is the To investigate a method of resolving quadratic equations, by means of trigonometrical tables. 1. Let r* =E pr= q be the equation A. proposed to be resolved. Suppose AB, and Bc, the perpendicu- N lar of base of a right angled triangle : otác - - p.A. EB = In #c Which are trigonometrical values of the two roots. 2. Let it pa - wo = q’ be the proposed equation. Then making AB = q, Ac = }p, it may be shown, similarly to the above, that cA + BC = BE and BD, are the two roots: whence the appropriate trigonometrical expressions may be readily deduced. th The precepts for all the cases may hence be laid down US — * 1. If the equation be of the form w” + pro- q* : - a 2 - The logarithmic application of these formulae is very simple, as will be manifest from the following Example. -- - - 7 Find the roots of the equation wo +. AT” tables of sines and tangents. sum – 10= log z = - 14685211 = log 294.1176. This value of a, viz. 294.1176, is nearly equal to #. To find if this be the exact root, take the arithmetical compliment of the last logarithm, viz. 0:5314379, and consider it as the logarithm of the denominator of a fraction whose numerator is unity: thus is the fraction found to be; exactly, which is manifestly equal to 5 - - 1695 T' The other root of the equation is equal to — ToTTG 5 339 + T = - His' ScholIUM. Cubic equations, as well as quadratics, are readily solved by means of trigonometrical tables. The formulae demonstrated and given by Cagnoli and Borda for this purpose, are as follow: 1. For cubics of the form ré + pt it q = 0. q 2. For cubics of the form w8 – pr:t q = 0. Then r = + cosec 2A. 2 väp. Here, if the value of sin B should exceed unity, B would be imaginary, and the equation would fall in what is called the irreducible case of cubics. In that case we their sum is equal to the third with a contrary sign, and that this third is the one which is produced from the first solution. . - - - * * In these solutions, the double signs in the value of r, relate to the double signs in the value of q. : "... ; *: - PRoRLEM V. To investigate series in which any circular arc shall be expressed in terms of its sine, or of its tangent. There are various methods of solving this problem: one of the simplest is by means of the expressions for the differentials of the sine and tangent of an arc, given in the preceding chapter. - A. Thus, from the expression 3 sin B = 2 B cos B, we have, * * - # sin B 3 sin B . . . .. – # - * – - * - cir. 2 * =# = w/(l #5= *sin a (1 – sin “B) *. Expanding the factor (1 — sino B) by the binomial theorem, the whole expression becomes - - 1-3 1.3.5 - - . I sin? — ci - in 6 + . 3B = 9 sin B(1 ++sin * + 31 sin" b + -ājāsin"B + &c.) - - - * . . " --The o of this expression is very easy: for the integral of 3B is B, that of 3 sin B x 1 is sin B, that |