¦ (log M + log B) = log MB = { log(2+2x B) B X = log √ (P2 + B2) = log H; as it ought to be. Scholium. The following formulæ, 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 b greater. Find x, ≈, &c. so, that tan x = √ sin z = a' log (a2b2) = log a + log sin y= log b + log tan y ✔ log (a - b) = [log (a + b) + log (a - b)]. (a2+b2)=loga + log sec u logb+ log cosecu. log (a + b) loga + log secxlog a +log 2 + log cos ty. log (ab)=loga + log cos zloga + log 2 + log sin by. log To investigate a method of resolving quadratic equations, by means of trigonometrical tables. x2+px 1. Let x px = q2 be the equation proposed to be resolved. Suppose AB, and BC, the perpendicular and base of a right angled triangle ABC, to be respectively equal to q and ap: then D B C CA = CD = CE = √(AB2 + BC2) = √ (q2 + {p2) = √(x2±px + p2) = x ±1p. E Consequently AB 24 Now = BC Р CA — 3p = DB or be. tan ctan 2E (Euc. iii. 20.) Also, DB: BA :: rad: tan D, or cot E, or cot c Therefore DB = EB Which are trigonometrical values of the two roots. 2. Let px-x2q2 be the proposed equation. Then making AB = 9, AC = p, it may be shown, similarly to the above, that CA BCBE and BD, are the two roots: whence the appropriate trigonometrical expressions may be readily deduced. The precepts for all the cases may hence be laid down thus: 1. If the equation be of the form x2 + px = q2; 2q : then will Р Make, as before, tan c = x= q tan c.... x = + q cot c. 3. For quadratics of the form x2 + px = Make sin c= (q2). X=- q tan c x= 4. For quadratics of the form x2 2q Make sin c = : then will Ρ x = + q tan c.... x = + q cot &c. In the last two cases, if The logarithmic application of these formulæ is very simple, as will be manifest from the following Example. 1695 Find the roots of the equation x2 + X= by tables of sines and tangents. 12716' log tan ¿c = 9·9061115=logtan 38°51′15′′ log q, as above sum 10=log x = 9.5624096 1-4685211 = log ⚫2941176. This value of x, viz. 2941176, is nearly equal to 5 17' 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 1 found to be exactly; which is manifestly equal to Cubic equations, as well as quadratics, are readily solved by means of trigonometrical tables. The formulæ demonstrated and given by Cagnoli and Borda for this purpose, are as follow: 1. For cubics of the form x3 + px ± q = 0. 2 √✓ p.... tan A = 3 tan B. cosec 2A. 2√3p. Here, if the value of sin B should exceed unity, в would be imaginary, and the equation would fall in what is called the irreducible case of cubics. In that case we xsin A.2 √ 3p. x = ± sin (60° A).2 √3p. If the value of sin в were 1, we should have B = 90°, tan A= 1; therefore a = 45°, and x = 2√p. But this would not be the only root. The second solution would give cosec 3A1: therefore a = x = sin 30°. 2 √ sp Here it is obvious that the first two = 90° ; and then ± √ 33p. +2√3p. roots are equal, that = 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 x, relate to the double signs in the value of q. PROBLEM 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. Thus, from the expression sin BB COŚ B, we Expanding the factor (1 sin2 B) by the binomial theorem, the whole expression becomes + &c.) The integration of this expression is very easy: for the integral of dB is в, that of sin B x 1 is sin в, that of sin Besin в is 2.3 = Suppose в to be 30°, then because sin 30° (ch. ii. pr. 5.) it results that |