AE AF × cotang. FAE; whence the angle AFE may be determined. And in the right-angled triangle DAF, AFX sin. AFD AD= rad. may be found; which line and the angle BAD determine the position of D. (94.) COR. 2. Also AB: AM :: rad. cos. BAC, R2: cos2. BAC. :. AB× AD : AM × AE :: and since AD AC, and AM × AE = AK× AL= P × [(P-BC) ABX AC: P. (P-BC) :: R2 cos2. + BAC, (95.) COR. 3. Also AE: ED :: R: tang. BAC, and AM: MB:: R: tang. BAC, .. AEX AM: ED× MB :: R: tang. + BAC, or P. (P-BC): (P-AB). (PAC) :: R: tang. (96.) Having given the sines and cosines of two arcs; to find the sine of an arc which is equal to their sum. Let AB, AC be the two arcs; take AD, AE their doubles; draw the diameter AF. Join AD, DF, AE, EF, ED. Then the chord AD=2. sin. AB (62) E B and DF=2 cos. AD=2 cos. AB (70), EF 2 cos. AC. Now AF× ED=AD × FE+AE × FD (Eucl. vi. D.) or 2 rad. x 2 sin. BC=2 sin. AB x 2 cos. AC + 2 sin. AC × 2 cos. AB, .. rad × sin. BC= sin. AB × cos. AC+sin. AC × cos. AB. (98.) COR. 2. If a=b, sin 2a = 2 sin. a x cos. a. sin. 4 a = (4 sin. a-8 sin3. a) . cos. a sin. 5 a 5 sin. a 20 sin.3 a + 16 sin3. a. PROP. XVII. (99.) Given the sines and cosines of two arcs; to find the sine of an arc which is equal to their difference. Let AB, AC be the two arcs; take AD, AE the doubles of them; draw the and AF× ED = AD × EF-AE× FD (Eucl. vi. D.) or 2 rad. × 2 sin. BC = 2 sin. AB × 2 cos. AC· [2 sin. AC × 2 cos. AB, .. rad. × sin. BC-sin. AB x cos. AC- sin. AC × cos. AB. (100.) COR. If rad. = 1, and AB = a, AC = b, sin. (a - b) sin. a x cos. b- sin. bx cos. a. PROP. XVIII. (101.) Given the sines and cosines of two arcs; to find the cosine of an arc which is equal to their sum. Let AB, AC be the two arcs. AD, AE the doubles of them. Take Draw the diameter AF, and make FG= AE. = B and AG = 2 cos. + FG=2 cos + AE = 2 cos. AC. AD and FG 2 sin. + FG = 2 sin. AC. And AF × DG = FD × AG—AD × FG (Eucl. vi. D.) or 2 rad. × 2 cos. BC = 2 cos. AB x 2 cos. AC - 2 sin. [AB × 2 sin. AC, .. rad. x cos. BC cos. AB x cos. AC-sin. AB x sin. AC. or cos. 2 α= 1—2 sin2. a. Also cos. 3 a = 4 cos3. a-3 cos. a cos. 4α = 8 cos1. a 8 cos2. a+1. cos. 5 a 16 cos5. a - 20 cos3. a +5 cos. a cos. 6a = 32 cos". α 48 cost. a +18 cos2. a -1. PROP. XIX. (104.) Given the sines and cosines of two arcs; to find the cosine of an arc which is equal to their differ ence. Let AB, AC be the two arcs, take AD, AE the doubles of them; draw the diameter AG, and make GF (on the opposite side) = AE. Join AD, AF, DF, DG, FG. The arc DGF is the supple F E ment of DE, which.. is equal to AD-AE= 2 AB 2 AC 2 BC, and .. DF = 2 cos. + DE = 2 cos. BC. DG 2 cos. AB (70) AF 2 cos. AC AD 2. sin. AB (62.) FG 2. sin. AC. Now AG - DF = DG × AF÷AD × GF (Eucl. vi. D.) or 2 rad. × 2 cos. BC= 2 cos. AB x 2 cos. AC+ 2 sin. AB [× 2 sin. AC. .. rad. × cos. BC=cos. AB × cos. AC+sin. AB × sin. AC, (105.) COR. If rad. = 1, AB = a, AC=b, cos. (a - b) =cos. α x cos. b+sin. a x sin. b. (106.) It may perhaps be objected that the preceding propositions are proved only when the arcs (a) and (b) or even (a+b) are less than quadrants. Assuming them to be proved within such a limit that (a) does not exceed a value a, and (b) a value ß, it may be proved by means of what has been shewn before, that the values of the sines and cosines of the sums are equally true, when (a) does not exceed 90° +a and (b) does not exceed B. 90o. For let a = 90° +m, ... m=a ~ 90°, which is less than Now sin. (90° +m+b) = sin. {90°— (m+b)} = cos. (m+b) = cos. m x cos. b- sin. m x sin. b sin. (90°—m) x cos. b-cos. (90°- m) × sin. b = sin. (180° — a) × cos. b − cos. (180° —a) × sin. b. = sin. a x cos. b+cos. a × sin. b. > Also cos. (90+m+b)=—cos. {90—(m+b)}. = = · sin. (m+b) sin. m x cos. b-cos. mx sin. b =-cos. (90-m) x cos. b- sin. (90 m) × sin. b = - cos. (180—a) × cos. b- sin. (180 − a) × sin. b =cos. α X cos. b- sin. a x sin. b. Hence these expressions which were demonstrated for (a) less than a and (b) less than ẞ, are also true when (a) does not exceed 90+a and (b) does not exceed ß. In the very same manner from the preceding, it might be proved that they are true, when (a) does not exceed 90+a and (b) 90+ß, and so on, i. e. they are true whatever be the values of (a) and (b) The same is equally applicable to the values of sin. (a - b) and cos. (a - b). (107.) Since sin. (30°+a) = sin. 30° x cos. a+cos. [30° x sin. a, and sin. (30° - a)=sin. 30° x cos. a-cos. 30° x sin. a, ... sin. (30°+a)+sin. (30° − a)=2 sin. 30° x cos. a=cos. a [(63). and sin. (30° + a)=cos. a-sin. (30°-a). If.. the sines and cosines of all arcs less than 30° be known, the sines of all arcs above 30° and less than 60° might be found by subtraction. (108.) Since sin. (60°+a) = sin. 60° x cos. a+cos. 60° [× sin. a, and sin. (60°- a)=sin. 60° x cos. a-cos. 60° x sin. a, ... sin. (60°+a) — sin. (60° - a) = 2 cos. 60° x sin. a=sin. a, and sin. (60°+a) = sin. a + sin. (60o sin. a + sin. (60° - a). If.. the sines of arcs less than 60° be known, the sines of arcs greater than 60° may be found, by addition. ЗА |