dications of an absurd supposition, they still serve as convenient instruments of investigation. Their utility must, therefore, be briefly shewn. The expression sin 2A + cos AR or 1, is resolvable into the imaginary factors, (cos A+ sin A-1). (cos A - - · sin A √ − 1) = 1. In like manner, introducing another arc в, and employing imaginary factors, we shall have (cos A + sin a ~ 1), (cos B+ sin B-1)=COS A COS B sin A sin B-1 (cos A Sin B + sin A cos B) = COS (A + B) + sin (A + B) √ 1, [by formula c]; also (cos a sin A 1). (cos B sin B✓ 1) = cos ( A + B) By a like calculation it will be seen that (cos A ± sin A√ 1). (cos в sin B √− 1) : (cos csin e √ 1) cos (A + B + C) sin (A + B + C) √1. It appears, therefore, that the multiplication of this class of quantities, is effected bysimply adding the arcs, a property analogous to the characteristic property of logarithms. sin (A+B) 1. 29. If the arcs A and B be supposed equal and two imaginary factors be taken, we shall have (cos Asin A 1) cos 2A sin 2A-1; if three factors be taken, the result, will be (cos Asin A-1)3 = cos 3A+ sin 3A −1, Whence, in general, (COS Asin A − 1 )” cos na sin na √ .1. 30. From the last equation we obtain sin na and sinna cos NA, 1 (cos A+ sin a ✅✅ — 1)” . 1= (cos Asin A-1)+ cos na, dividing the sum of these by 2-1, there results the following expression for the sine of any multiple arc; viz. D 31. These quantities may readily be expanded into series, by means of the binomial theorem; in which case all the terms affected with imaginaries will be annihilated, and there will be obtained the two following general series, one for the sines the other for the cosines of multiple arcs. n (n 1) (n-2) 1.2.3 3. A .... (x.) cosn 6 A sin A+ &c. n (n - 1) (n-2) (n − 3) n (n − 1) (n − 2) (n − 3) (n − 4) (n — 5) 32. If the arc A be supposed indefinitely small, then will sin A = A, cos A = 1, and that the arc na may become an assignable quantity, n itself must be regarded as indefinitely great; in that case unity will (practically) vanish in the terms n- 1, n 2, &c. so that n (n - 1) will become n2, n (n 1) (n-2) = n3, &c. Hence, if we put na = c, we shall have (since sin a = a, and sin 3A = &c. while cos A, COS2 A = n A, cos 3A, will each 1. The two last formulæ will, therefore, be reduced to the following, by means of which it will be easy to express the sine and cosine of any arc, in parts of that arc, or in decimals of the radius, that is, to calculate the natural sine and cosine of such These series will converge rapidly when the arc is small.* 33. From expressions, such as the preceding, for the sines, tangents, &c. of the sums, differences, and products of two arcs, or angles, it would be easy to pass to those for three, four, or more arcs. But as the properties which might thus be developed, however curious and elegant, are comparatively of little utility; we shall not present them here, but confine our investigation, either to the angles of a plane triangle, or to those which have an obvious relation to them. A, B, and C, being the three angles of a plane triangle, since c = 180° — (A + B), tan Atan B+ tan ctan A tan в tan c. (4.) Dividing this equation by the whole of the first member, and substituting for the products of the tangents divided by their sum, their corresponding values in cotangents (from equa. s), there will result, 1 cot A cot B + cot B cot c + cot a cot c.... (5.) If the sum of A, B, and C, instead of being 180°, were 360°, the same formulæ would result, as is evident from the consideration, that in that case also we should have tan c tan (A + B).† It may, farther, be readily seen that the same expression is applicable, so long as A+B+C = n. 180° or 2n.90°. This branch of the subject is pursued, with great elegance, to a considerable extent, by Euler, in his valuable work, Analysis Infinitorum. 2609, + The property restricted to the case when A + B + C = was announced and demonstrated by Dr. Maskelyne in the Philosophical Transactions for 1808: in the case of the triangle it had been previously demonstrated by Caguoli. 34. Suppose A + B + C = 90°, then tan c = cot 1-tan A tan B (by equa. o.) or, tan-A tan c+tan B tan c = 1-tan A tan B (by mul.) tan A tan B+tan-A tan c+tan B tan c (by trans.): or, 1 and cot A cot B cot ccot A + cot B + cot c..........(6.) → And this formula will obtain, in like manner, so long as A + B + C = (2n + 1) 90o. 35. Substituting for tan, its value preceding, it becomes sin -- COS in equa (4) sin c sin A sin sin c + COS A COS B whence, taking away the denominators, there results, sin A COS B COS C. + sin B COS A COS.C + sin C COS A COS B = sin A Sin B sin c (7.).. 36. By adding 3 sin A sin в sin c to both sides of equa. (7) we have 4 sin A sin B sin c. sin a cos B. COS C + sin A sin B sin c. + sin B COS A COS C + sin B sin A sin C + sin ccos A COS B + sin c sin a sin B sin A COS (CB) + sin B COS (CA) + sin c cos (B — A) =sin (A+C-B) + sin (A-C+B) + sin (B+C-A), +sin(B-C+A) + (C+B-A)+sin(c-B+A) = sin (A+C-B) + sin(4+B — C) + sin (B+C—A). sin 2B + sin 2c + sin 2A.... (8.) = sin Let A A+ 90°, B = B + 90°, c = c + 90°, then we shall have 4 COS A' COS B' cos.c′ = cos 2a + cos 2B + cos 2€,. (9.) The formula 7, 8, 9, are due to M. Mello; the last (2n+1) 90° applies to the case where A' + B' + C′ = 37. We may now add all that is requisite in the solution of rectilineal triangles: and shall first take the case where, instead of having two sides a, b, and their in cluded angle c given, the logarithms of those sides are given; as frequently happens in geodesic operations, and in astronomical tables for the distances of the planets from the sun. Here if a and b are regarded as the sides of a right angled triangle, in which a denotes the angle opposite to the side a, we shall have τα tana. But, since a is supposed greater than 6, this angle will be greater than half a right angle, or will be measured by an arc greater than 45°. Hence, because tam (« —45°) — ·· and because tan 45°R=1, we have tan (a · 45°) = (~~~1) +1+ Consequently, from equa. (3) -b tan (A —B) = a + b tan (A + B) tana- tan 45° 1+tan a tan tan † (A — B). whence cot &c tan (AB) cotic tan (45°).... (10.) Hence results the following practical rule: Subtract the less from the greater of the given logs, the remainder will be the log. tangent of an angle: take 45° from that angle, and to the log. tangent of the remainder add the log. cotangent of half the given angle, the sum will be the log. tangent of half the difference of the other two angles of the plane triangle. 38. Let us next return to the equation (1) in art. 11 of this chapter; namely, a = b eos c + C COS B ba eos c + c cós A ca éos Bb Cos A The first of these equations being multiplied by a, the second by b, the third by c, and each of the equations thus obtained, being taken from the sum of the other two, there will arise |