These results should be committed to memory. The sine of the difference of any two angles equals the sine of the first into the cosine of the second minus the cosine of the first into the sine of the second. The cosine of the difference of any two angles equals the cosine of the first into the cosine of the second plus the sine of the first into the sine of the second. The tangent of the difference of any two angles equals the tangent of the first minus the tangent of the second divided by I plus the tangent of the first into the tangent of the second. and by substituting these values for A and B we get = 2 sin. X sin. These results are most important, and should be thoroughly understood and committed to memory, as follows The sum of the sines of two angles equals twice the sine of half their sum into the cosine of half their difference. The difference of the sines of two angles equals twice the cosine of half their sum into the sine of half their difference. The sum of the cosines of two angles equals twice the cosine of halt their sum into the cosine of half their difference. The difference of the cosines of two angles equals twice the sine of hali the sum into the sine of half their difference, reversing the order of the terms. In addition to the above the following results, which are easily proved. should be remembered. 2 The cosine of any angle equals the cos.2 of its half, minus sin. of its half, i.e. The cosine of any angle equals 1 - 2 sin.2 of its half, i.e. The cosine of any angle equals twice the cos. of its half, minus I, i.e. A+ B 2 A+ B sin. X cos. A B 2 A B 2 Let A B C, Fig. 6, be a triangle having the angles acute; from C draw CD perpendicular to A B. Let C A = b and C B = a. Case II.-When one of the angles is obtuse, as B. Let A B C, Fig. 7, be a triangle having angle B obtuse; produce A B to D and draw DC perpendicular to A D. PROOF OF RULE OF COSINES Case I. When the angle is acute. Let A B C, Fig. 8, be a triangle having A acute; from B draw B D perpendicular to A C. Let A B C, Fig. 9, be an obtuse-angled triangle having the angle A obtuse. From B drop a perpendicular on C A produced to D. = CA2 + A B2 + 2 CA. BA. (— cos. A); because A D = = CA2 + A B2-2 CA. BA. cos. A Logarithmic formula for sine of half an angle of a triangle in terms of the sides A 2 = = 2 b2 + c2 — a2 2 b c b2 c2 + a2 2bc a2 — (b2 + c2 - 2 b c) a2 ·(bc) 2 2bc (ab+c) (a + b—c) a + b + c = 2 s a − b + c = 2 s — 2 b = 2 S 2 c = 2 (s A 2 = sin. 2 2 Isin. A = 2 (sb) 2 (sc) (s—b) (s — c) bc b) (s bc 1 {log. (s — b) + log. (s — c) + 20 — (log. b + log. c)} |