CHAPTER II. 7. To find the sine and cosine of the sum and difference of two angles. = Let the angle FAC=A, and the angle FAP B, then PAC=A+B, draw PF perpendicular to AF, and from P and F let fall the perpendiculars PD and FE on AC, and draw FG parallel to AC. Now, because GF is parallel to AC, the alternate angles, AFG and FAC are equal, A B P D E F but FAC-A, therefore AFG = A; and since AFP is a right angle, the angle PFG is the complement of AFG; but FGP is a right angle, therefore FPG is the complement of PFG, that is, the angles FPG and AFG are the complements of the same angle (PFG), they are therefore equal; but AFG has been proved equal to A, hence FPG = A. sin (A + B) = sin A cos B + cos A sin B. = AD AE-DE AE GF = AP AP AP AP' (because GF = DE) any To find the sine and cosine of A- B. Let the angle DAF = A, and the angle PAFB, then ▲ PAD = A – B : from point P draw the perpendiculars PF and PD on AF and AD, also draw PG parallel to AD, then 4 PFG = the complement of A AFE, and therefore A, because AFG is = F G E D the complement of A; and because GD is a rectangle, ED PG, and PD = GE. = 8. Now, since sin (A + B) = sin A cos B + cos A sin B, if we make B = A, then sin (A+A) = sin 2 A = sin A cos A+ cos A sin A = 2 sin Acos A ; also, since cos (A+B) = cos A cos B- sin A sin B; make B but since cos2 A = 1 − sin3 A, we have also cos 2 A = 1 - sin3 A – sin2 A − 1 − 2 sin3 A ; = be = also sin2 A = 1 - cos2 A ; cos3 A - (1 - cos2 A) = 2 cos2 A − 1. = A; sin' A ; ... cos 2 A Since A may any angle, we have the sine of any angle equal to twice the sine of half that angle multipled by the cosine of half that angle; and the cosine of any angle is equal to the square of the cosine of half that angle minus the sine square of half that angle. For the sines we have By putting 2 A for A, and A for B in the sin (A + B), we have sin (2A + A) = sin 2 A cos A + cos 2 A sin A; but sin 2 A = 2 sin A cos A, and cos 2 A = 1 − 2 sin2 A sin (2 A + A) = 2 sin A cos A cos A + (1 − 2 sin2 A) sin A, = 2 sin A cos2 A + sin A - 2 sin3 A, = 2 sin A (1 − sin3 A) + sin A − 2 sin3 A, cos 3 A = cos (2A + A) = cos 2 À cos A − sin 2 A sin A, 2 sin A cos A sin A, 2 cos A sin A, 2 cos A (1 − cos2 A), 2 cos A +2 cos3 A, In the same way we may find the sin 4A, &c., but it will be better to express them in general terms, as in the next Article. 10. Now, since sin (A + B) = sin A cos B + cos A sin B, by addition we have sin (A+B) + sin (AB) = 2 sin A cos B...(1). By subtraction, sin (A + B) – sin (A – B) = 2 cos A sin B .........(2). By multiplication, sin (A + B) sin (A – B) = sin2 A cos2 B - cos2 A sin3 B sin A (1 sin' B)- sin B (1-sin3 A) = = sin3 A – sin3 B = (sin A + sin B) (sin A - sin B)...(3). Also, cos (A + B) = cos A cos B – sin A sin B, cos (AB)= cos A cos B + sin A sin B. By addition, cos (A + B) + cos (A – B) = 2 cos A cos B............(4). By subtraction, cos (AB)-cos (A + B) = 2 sin A sin B ............(5). By multiplication, cos (A + B) cos (A - B) = cos3 A cos2 B - sin2 A sin2 B = (cos A + sin B) (cos A - sin B) ....... sin (A + B) + sin (A – B) = 2 sin A cos B, for B put a, and for A put na; then, A+B = na + a = (n + 1) a, and A – B = na· ·(6). - a = (n − 1)a, sin (n + 1)a + sin (n − 1)a = 2 sin na cos a, cos (n + 1)a + cos (n − 1) a = 2 cos na cos a. By giving to n different values, the sine or cosine of any multiple arc may be found. 11. Now, since half the sum of any two quantities added to half their difference gives the greater, and half the difference subtracted from half the sum gives the less, we have Putting A = ↓ (A + B) + ¦ (A − B), (A + B) for A, and (A + B) and cos (A + B); = 1 (A – B). (A – B) for B, in sin · sin 1 (A + B) cos 1⁄2 (A − B) + cos 1⁄2 (A + B) sin § (A – B)...(1). In the same manner, sin B=sin(A+B) cos (A–B)—cos (A+B) sin (A−B).....(2). By adding (1) and (2), sin A+ sin B = 2 sin (A + B) cos (A – B) ..(3). By subtracting (2) from (1), sin A-sin B = 2 sin (4B) cos (A + B) ..... ··(4). cos A = cos {(A + B) + (A − B)} = cos(A+B) cos (A-B) - sin & (A + B) sin (A– B)...(5). cos B = cos {1⁄2 (A + B) − ¦ (A − B)} = cos (A+B) cos † (A − B) + sin § (A + B) sin § (A−B)...(6). |