142 



THE POPULAR EDUCATOR. 



These cases will probably convince the student that (33) and 

 (34) hold good for all values of A and B, as can, indeed, be 

 proved separately, in the same way, for every value. As prac- 

 tice, the student should prove the following cases : 



EXEECISE 3. 



1. Prove (33), where A is an angle in the third, and B an angle in 

 the first quadrant, but where A + B reaches to the fourth quadrant. 



2. Prove the same where both A and A + B are in the third quad- 

 rant (which, of course, implies that B is less than a right angle). 



3. Prove (.34), when A exceeds 180, but is less than 270, and 

 when B exceeds 90, but is less than 18U. Construct the figure 

 on the supposition that A is so near 270, and B so much less than 

 180, that A B falls in the second quadrant. Also constluct it so 

 that A - B sh;ill be less than 90. 



4. Prove (34), when A is an angle in the fifth quadrant, and when 

 B = 180. In this exairple AO B must, of course, be drawn as an 

 angle in the first quadrant, and since BOC = 180, BO and oc ,are in 

 line with each other. P Q is therefore the only other line in the 

 construction before given which it is possible to draw. A B = the 

 (trigonometrical) angle A D c iu the third quadrant. 



Then sin. ( A B) = P - Q = sin. A ; 



o P 



since P OQ = A o B, and sin. A is naturally positive. 

 This agrees with (34), where, if wa substitute the values of sin. and 

 cos., 180, as given in Secc. VIII., we get- 



Sin. (A B) == (sin. A x 1) (cos. A x 0) = sin. A. 



5. Prove (34), where A = 180 and B exceeds 90. 



(35) and (36) can also be proved geometrically. (35) can, 

 however, be proved more shortly, thus 



Since sin. A = cos. (90 - A), and vice versd; 

 Cos. (A + B) = sin. ^90 - (A + B)) = sin. ((90 - A) - B\ 



Whence, by (34), 



Cos. (A. + B) = sin. (90 A) cos. B - cos. (90 - A) sin. B 

 = cos. A cos. B. sin. A sin. B. 



To prove (36) : 

 Cos. (A - B) = sin. (90" - (A - B)) = sin. ((90 A) + B) 



= sin. (90 - A) cos. B + cos. (90 - A) sin. B 



= cos. A cos. B + sin. A sin. B. 



(34) can also be derived from (33) by substituting - B for B 

 in (33). The student should work this out, remembering that 

 sin. B = - sin. B, but cos. - B = cos. B. 



XII. Formulae for the Sum and Difference of the Sines and 

 Cosines of the Sum and Difference of two Angles. By adding 

 together (33) and (34), we obtain 



Sin. (A + B) + sin. (A - B) = 2 sin. A cos. B ...... (37) 



By subtracting (34) from (33) 



Sin. (A + B) - sin. (A - B) = 2 cos. A sin. B ...... (38) 



By adding (35) and (36) 



Cos. (A + B) + cos. (A B) = 2 cos. A cos. B ...... (39) 



By subtracting (36) from (35) 



Cos. (A + B) - cos. (A - B) = - 2 sin. A sin. B... (40) 



XIII. Formulae for the Sum and Difference of the Sines and 

 Cosines of two Angles : 



.-. by (33), sin. A= sin.^+lcos. 

 2 



Similarly, sin. B = sin. 



+ co s . A + B ,in. ^_ 



E 



A ~ B 

 2 



. A + 3 A-B A + B . A-B 



= sin. - cos cos. . sm. . 



22 22 



Adding these results together, we get 



Sin. A + sin. B = 2 sin. A +-? . cos. ^"-J? 



(41) 



Or, subtracting one from the other 



Sin. A - sin. B = 2 cos. A --t? . sin. A ~ B (42) 



2 2 



Similarly, by adding and subtracting like expressions for 

 oos. A and cos. B, we get 



Cos. A + cos. B = 2 cos. A .+ ? . 

 2 



Cos. A - cos. B = - 2 sin. A + B . sin. 



Aj-_B 



2 

 A-B 



(44> 



XIV. Relations between Sines, Cosines, and Tangents of tw>t 

 Angles. Dividing (33) by (35), we have 



T ( A _l_ "R\ S * n ' A COS- -^ ~t~ COS< A S * n- -^ 



lan. (A. + D) . - ; - . 



cos. A cos. B sin. A sin. B 



Dividing both numerator and denominator on the right-hand 

 side by cos. A cos. B, we have 



Tan. (A+B) = 



sin. A , sin. B 

 cos. A cos. B 



1 - 



sin. A 

 cos. A 



sin. B 

 cos. B 



/ A , T>\ tan. A + tan. B 



.-. tan. (A + B) = - L . 



1 tan. A tan. B 



(45) 



Similarly, dividing (34) by (36), and again dividing the nume- 

 rator and denominator by cos. A cos. B, we obtain - 



Tan. (A-B) - tan. A - tan. B 46> 



1 + tan. A tan. B 

 Again, dividing (41) by (42), we obtain 



Sin. A + sin. B _ 2 sin. \ (A + B) cos. (A - B) 



Sin. AT- sin. B ~~ 2 ' cos. (A + B) ' sin. i (A - B) 



= tan. HA + B) cot. * (A - B) ; 



.'. since cot. $ (A - B) = , 



tan. (A - B) 



sin. A + sin. B _ tan. ^ (A + B) , . _, 



sin. A - sin. B tan. ^ (A - B) 



Or, the sum of the sines of two angles is to the difference of their 

 sines as the tangent of half their sum is to the tangent of half 

 their difference. 



Similarly, by dividing (43) by (44) 



Cos J _Ajf_ooa.JB _ cot. i (A + B) 

 Cos. A - cos. B tan. ^ (A - B) 



(4S 



XV. Formulae for the Ratios of the Sum of three Angles may 

 be obtained simply by splitting up the three into two, which 

 can then be dealt with by formulas already given ; thus 



Sin. (A + B + C) = sin. (A + (B + C)) 



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

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



+ cos. B sin. C). 



Whence, by a slight change in order 



Sin. (A+B + C) = sin. A cos. B cos. C + sin. B cos. A cos. C ") , ,Q, 

 + sin. C cos. A cos. B - sin. A sin. B sin. C $ ' 



By similar reasoning 



Cos. (A + B + C) = cos. A cos. B cos. C cos. A sin. B> ,KQV 

 sin. C - cos. B sin. A sin. C - cos. C sin. A sin. Bj 



Dividing (49) by (50), and the numerator and denominator of 

 the fraction thus obtained by cos. A cos. B cos. C, we obtain 



Tan. (A + B + C) = } 



tan. A + tan. B + tan. C tan. A tan. B tan. C > ... (51) 



1 - tan. A tan. B - tan. A tan. C - tan. B tan. C J 



XVI. Formulae for the Ratios of the Multiples of an Angle. 

 Substituting A for B in (33), we have 



Sin. (A + A) = sin. A cos. A + cos. A sin. A ; 



.'. sin. 2A = 2 sin. A cos. A (52) 



Similarly, by (35), cos. (A+A) = cos. A cos. A - sin. A sin. A ;. 



.-. cos. 2A = cos. 2 A - sin. 2 A (53> 



By (7), 1 = sin. 2 A + cos. 2 A ; adding this to (53) 



Cos. 2A = 2 cos. 2 A - 1 (54) 



Subtracting (7) from (53) 



Cos. 2A = 1 - 2 sin. 2 A (55> 



Again, substituting A for B in (45), we have 

 2 tan. A 



Tan. 2A = 



1 - tan. 2 A 



(56) 



