Ex. 3. Given axbarc; to find x. Taking the log. of each side of the equation, In calculating by logarithms it appears, therefore, that multiplication is effected by addition, division by subtraction, the raising of powers by multiplication, and the extraction of roots by a process of division. The use of the tables may be learnt from the explanations given in the prefaces to such books. 19. Since log 10,= n, .. log (N × 10") = log N + log 10” = log N + n ; from this it follows that if log 3·754=·57449, then log 375-4 or 3·754 × 102 = log 3·754 + log 102 = ·57449 + 2 =2·57449. And so on to other cases. from this it follows that if log 5.621=7498, then log ⚫05621 =7498—2, or as this is usually written 2.7498. And so on to other cases. 20. To avoid minus characteristics in the logarithm tables of sines and cosines, the radius is taken 1010. If sin a be the symbol for the sine of a when the radius of the circle is unity, and Sin A the symbol for the sine of the same arc when the radius is 1010, then by Art. 10., we have, Sin A 1010 sin A= ;.. log sin a log Sin A-10. 21. To calculate a from the equation sin A = Art. 11.) Here ab sin a; then by Art. 16., where any found. -= 10...... (1). sin B с sin c -log Sin c (2); three terms being given, the remaining one may be As an illustration of this method of calculation let us take a few examples. Ex. 1. In the right-angled triangle ABC (see fig. Art. 11.), AC= 363, and ▲ a = 63° 25′; required BC. Ex. 2. In the triangle ABC (Art. 11.), given c=50, ≤ a = 74° 14′, ▲ B = 49° 23', and .. 4c 56° 23′; to find b. = Here from eq. (2), Art. 21., we have, log blog c + log Sin в — log Sin c As further exercises the student may solve, by the method of logarithms, all the examples given in Arts. 11. 12. and 13. 22. To trace the signs of sin a, cos ▲, &c. as the arc ▲ increases from 0° to 360°. An inspection of this figure will show, 1. When the arc CD is less than 90°, the sine and cosine are both positive. 2. When the arc CE is greater than 90°, the sine, EI, is positive; but the cosine, AI, is negative. 3. When the arc CQF is greater than 180° or K E I H D 2 right angles, the sine, FI, is negative; and the cosine, AI, is also negative. 4. When the arc CQHG is greater than 270° or 3 right angles, the sine, GB, is negative; but the cosine, AB, is positive. 5. Let 2 be put for the whole circumference of the circle or 360°, and let ▲ be put for the degrees in any arc CD, then if we add 2 π to A it will obviously bring us to the same point D in the circle; or generally if we add any number of times 2 π to A, it will always bring us to the same point in the circle; and this will be true whether a be taken below the diameter CK or above it, that is, whether A be positive or negative. Hence we have, sin A = sin ( n × 2π+A), where the sign will be positive or negative according as a is positive or negative, or according as the point D is in the semicircle CQK or CHK. In like manner cos a = cos (n × 2ñ +▲), where the sign will be positive or negative according as the point D is in the semicircle HCQ or QKH. 6. When A=0 the sine, DB, evidently becomes O, and the cosine, AB, becomes aC or 1; when a= 90° the sine DB, becomes QA or 1, and the cosine, AB, becomes 0; when A = 180° the sine again becomes O, and the cosine becomes-1; when A= 270° the sine becomes-1, and the cosine becomes 0; when A= 360° the sine becomes O, and the cosine becomes 1. Expressing these results analytically, we have, sin 0=0, cos 0= 1, sin Π 2 Π = 1, cos = 0, sin == =0, 2 cos 2π=1. It is also evident that if any of these arcs be increased by 2 or any multiple of it, the result will not be altered. 7. If the student thoroughly understands the nature of these results, he will find no difficulty in tracing the signs of any trigonometrical expression which may hereafter occur. For example, let it be required to find the sign of the tangent of an arc A, greater than 90° and less than 180°. In this case, sin A is positive, while cos a is negative, COS (A + B) sin A. COS B+ cos A. sin B ... (1) COS A. COS B - sin A. sin B... (2) sin B (3) = sin A. cos B-COS A. sin (AB): COS (AB) = = COS A COS B + sin A. Let GADA, and DAB=B, then LGAB A+ B. From B, any point in AB, let fall BC and BD perpendiculars on AG and AD; and from D, let fall DG and DF perpendiculars, on AG and BC. Then CF = GD, FD=CG, and because DF is parallel to AG, LADF=A, .. ▲ DBF = 90° — / BDF = LADFA. Now because BACA + B, sin B CAB=A - B. Let L CAD = A, and BAD = B, then From B, any point in AB, let fall BC and BD perpendiculars on AC and AD; from D let fall DG perpendicular to AC; and from в let fall Bг perpendicular to Then BFCG, BC FG, and BDF = 90° - ADG=/ GAD=A. Now be D G. = A cause These four expressions constitute the fundamental formulæ in analytical trigonometry, from which a great variety of useful properties may be readily derived. The values of A and B being perfectly general, we may assign to them any particular values we please. 24. To prove that sin 2 A=2 sin a. cos a. In formula (1) let B = A, then A + B = 2 A, and .. sin 2 Asin A. cos a + cos a. sin a = 2 sin a. cos A. 25. To prove that cos 2 A= cos2 A-sin2 A, &c. In formula (2) let B = A, then we have, Cos 2 A = COS A . COS A - sin A. sin a cos2 A-sin2 A. In this expression put first 1 - cos2 A for sin2 A, and afterwards 1 - sin2 A for cos2 a (Art. 5.) then, cos 2 A= Cos2 A Cos 2 A = 1 - (1 — cos2 a) = 2 cos2 a 1 26. In the formulae of the last two articles, let 2 A = a, and |