SINES, ETC., or TWO ANGLES.] M ATHEM ATICS. TRIGON OMETR Y. 



617 



by observing that the sine of the sum of two angles 

 is the sum of the product of- the sine of the first angle 

 and the cosine of the second, and of the product of the 

 sine of the second angle and the cosine of the first ; while 

 the sine of the difference of the angles is the former pro- 

 duct minus the latter. The cosine of the sum of two 

 angles is the product of the cosines of those angles minus 

 the product of their sines ; and the cosine of the diti'er- 

 ence of two angles is the sum of the product of the 

 cosines and of the product of the sines. 



16. To prove the Formula 

 Sin. (A -f B) = sin. A cos. B + sin. B cos. A. 



Fig. 12. 



Let AOB be the 

 angle A, BOO the 

 angleB ; .'. AOC 

 is the angle A-f-B. 

 In O C take any 

 point P, and from 

 P let fall PN.PM, 

 perpendiculars OE 

 OAand OB, and 

 from M let fall 

 M Q and M R, per- 

 . pendiculars on PN 

 and OR. 



Then 



Now 



QN , PQ 



MR 

 OP 

 MR 



PQ 

 OP 



MR 

 OM 



sin. A and 



OP 

 OM 

 ' O P 

 OM 



PQ 

 " PM 



PM 

 OP 



= cos. B 



ON u O P 



And since PMO is a right angle, QPM=QMO = MOR, 

 . PQ OR PM . ,, 



- 



'. sin. (A + B) = sin. A. cos. B + sin. B, cos. A. 



17. To prove the Formula 

 Cos. (A + B) = cos. A. cos. B sin. A, sin. B. 

 For as before, 



Cos.CA+B^ ON = OR-QM OR_QM 

 ' OP OP OP OP 



OR OM_QM PM 

 " OM ' OP PM ' OP 

 OR . OM 



QM MR 

 = = 



PM 



= sin. B ; 



.'. cos. (A + B) = cos. A. cos. B sin. A. sin. B. 



18. To prove the Formula 



Sin.(A-B) = sin. 

 A. cos. B sin. B. 

 cos. A. 



Let A O B = A. 

 B O C = B. Take 

 in O C any point P ; 

 from P draw PM 

 and P R perpen- 

 dicular to OB and 

 O A. Draw M N 

 and M Q perpen- 

 dicular to OA and 

 M N. Then A O 

 = A-B. 



MR_QP 

 OP OP 



^. - cos. B. 



Again 



BOA = A. 



PQ QR 

 MP = OM = 

 .'. sin. (A B) = sin. A. cos. B - sin. B. cos A, 



, PM . 

 A -op =smeB - 



19. To prove the Formula 

 Cos. (A B) = cos. A. cos. B + sin. A. sin. B. 



ON OR MQ OR OM , MQ MP 

 cos. (A-B)= Qp- = oF O~~P = OM'^^ "*""** ^^ 



OR , OM 1 



But 0jj=cos. A, o^p = cos. B. j 



sin. A and 



MP 

 OP 



sin. B. 



.'. cos. (A B) = cos. A. cos. B + sin. A. sin. B. 



20. To extend the above proofs in special cases. 



The proofs above given are clearly limited to the cases 

 in which A, B, and A + B or A B, are each less 

 than 90. They admit of extension to any case what- 

 ever. Thus 



If A is > 180 < 270, B > 90 < 180, and A-B 

 > 90 < 180. 



To show that 



Sin. (A B)=sin. A cos. B sin. B cos. A. 



In this case the figure 

 will be the following : 



A O B is the angle A, 

 measured as indicated 

 by the dotted circle. 

 B O C is the angle B. 

 C O A is the angle 

 A-B. 



From any point P in I 

 O C draw PN, PM per- 

 pendicular to O A and 

 O B produced, and from 

 M draw M Q, MR per- 

 pendicular P N and O A. 



Then 



Fig. 14. 



' PO PO^PO PO 



, MR PQ PM MR POM 

 r PO = PM ' PO + MO ' PO 



Now 1^ cos. M P Q = cos. A M=cos. (A - 180) 

 = cos. (180- A) = -cos. A. 



=sin. P O M=sin. (180-B)=sin. B. 



MR 

 Similarly ^-j^- =sin. M O A= sin. A 



OM 

 PM 



=cos. POM=-cos. B; 



.'. Sin. (A-B)=sin. A. cos. B-sin. B. cos. A. 



In the same case, 



,. _. ON NR OR 

 cos. (A B) = = 



QM 

 OP ~ 



OR 

 OP 



iVl(j Ar*. i *-iv vxi'A 



^"^^ ^^ TrfSf ' f\v 



OP 



OR OM 

 CM' OP' 



Now = sin. MPQ: 

 M" 



MP 

 MP" OP 



sin. MOA = sin. A, 



^ = sin. POM - sin. FOB 



sin. B. 



PJ? = cos. MOA : 

 OM 



cos. A. 



: cos. B. 



. cos. POM, 



.'. cos. (A B) ** cos. Acos. B + sin. Asin. B. 



sin. A cos. B + 



Again, to show that sin. (A + B) 

 in. B cos. A. 



When A + B 7 270 ^ 300 

 B 7 180 Z. 270. 



TOL. J. 



A 7 90 ^ 180 

 4 K 



