MATHEMATICS. TRIGONOMETRY. 



[DI MOIVUB'S THBO&nt 



Hence, in the limit when equals aero, 



,. * , 1 







Cor. Under all circumstance* 



tan. sin. . 1 



Now when 



.,. i> 



Q. E.D. 



-1. 



Hence, in the limiting case, when i 

 tan. 

 



-L 



(48). /fc Moivn't Theorem. 



(a). To prove that (cos. + J^lsw^O) x (cos. + 

 ^Tigin. 0)- cos. (0 + 0)+ ^/ lain. (0+0). 

 For by actual multiplication 

 (cos. 6 + v/ 1 sin. 0) (cos. + ^/ 1 sin. 0) = 

 oos. cos. + V 1 " e s - + \/ 1 o 08 - ggin - 

 sin. sin. = cos. cos. sin. 0sin. + \/ 1 

 {sin. 0cos.0+cos.0sin.0j = cos. (0 + 0) + -J 1 

 sin. (0 + 0). 



Hence 

 (cos. + -x/ 1 sin. 0) (cos. + ,/ 1 sin. 0) (cos. V' 



+ J^l sin. YT) - jcos. (0 + 0) + s/^Tsin. (0 + 0) j 



(cos. y+v/^sin. -f)=.cos. (0 + 0+^)+^"^! 



sin. (6 + + YO- 



And generally if we had n angles X . . a ... 0., we should 

 have 



cos. 



Tsin. 



Isin. 0, | j 



^Tl sin. 0. | -cos. (0 t + 0, + ..... 0, ) 



^1 sin. (0, + a '+ ... 0, ) ...... (40). 



(6) To show that (cos. 0+v/ 1 sin. fl)" = cos. n + 

 ,J 1 gin. n for all integral and positive values of n. 



In equation (40), suppose r 0, ... to bo each equal 

 to one another and to 0. 



Then 0, + 0, ... + 0. = n 0. 



And 



(cos. t + V 1 sin. 0,) (cos. 2 + v/ 1 sin. 2 ) 

 ____ (cos. 0. + V- 1 sin. B ) = (cos. + V~l sin. 6) 



(cos. + ,/ - 1 sin. 0) ---- (cos. + V - Isin. 0) - 



(cos. 8+ V~ Isin. 0),. 



And cos. (0,+0 1! + ... + 0.) + s/ lain. (0, + 2 + 

 .. + 0.) _ 



- cos. n + j^/ 1 sin. n 

 Hence 

 (COB. + V 1 sin. 0)" = cos. n + J 1 sin. n 0. 



(c). To prove the same theorem when n is a negative 

 whole number. 



For by multiplication, (cos. n 0+ V 1 sin. n0), 

 (cos. n0 ^^~1 sin, n 0) = cos. 2 n0 + sin. 2 n0 = 1. 

 nowl - ( cog - e + V^Jsin.0)' 

 (cos. 0+ s/ 1 sin. 0) 



(eos. fl + V i s m- 0), " (cos. + N / 1 sin. 0) 



(cos. n + V 1 sin, n 0), (cos. 6 + J Isin. 0)-" 



.'. (oos. n v' 1 sin. n 0), (cos. 9 + ^ 1 

 siu. 0)-" = 



(OOB. n0+ v/^! Kin. n0), (cos. n0 V lgin.nfl); 



.'. (cos. + J^l sin. 0) -" = cos. n V 1 



sin. n 0) 



- cos. ( n0)+ j^-1 sin. ( n 0). 

 Since cos. ( nO) cos. n 



sin. ( n 0) = sin. n. 0. 

 Hence the theorem is true when n is negative. 



^1 sin. 0) 



( J). To prove the same theorem when n is fractional. 



Now (oos. + V^lsin-^Y- oos.p + ^^ 



sin. p 0. 



(cos. + v' 1 sin. e) _ 

 .'. oos.^+^/^^1 sin. *** xX(co 



= (cos. 6+ V^^l sin. 0) f 

 and hence, in all cases 



(cos. + J ] sin. 0; " = cos. n + V 1 



sin. nO ..... (41). 



Whether n be positive or negative, integral or fraction ; 

 which theorem is called De Moivre's theorem. 



(49). To express sin. m and cos. m Q in term of powers 



of tin. and cos. 0. 

 Since 



(cos. m + V 1 sin. m 0)j= (cos + V 1 <"* #)" 



cos. "-0sin.0 



+ 



tn. 



= COS."0-^- 



-l. m-2. m-3 

 1.9.8.4 



I0ain0 



&o. 



1.2.3 



cos. *- J 0sin. s + 

 Hence equating possible and impossible quantities, 



cos. m = cos "0 





+ '.">-l.; r 2.m-3 co3m _ tsin4g _ <fcc (42) 



1.2.3.4 



and 



Sin. m = m cos." - ' sin. 



m . TO 1 . tn 2 

 1 . 2 . S~~ 



cos. "-'0 sin. ' + . . . (43)* 



(50). To express sin. 0, and cos. 0, in terms of 0. 



Let m = 0, and .". 0=> 

 m. 



Hence in the expression of the last article for cos. m 

 0, we have 



f.JA 



/ rt\- V "*// 0\"~*/ - s^\_' 



- r *-) ~ "^ r -) ( -r 



1.2.3.4 



&o. 



] 



* We assume in this article that if o + / 1. 4 = A + 

 f/ 1 B. where a. ft. A. B. are real. Then a = A and B = ft. 

 To prove this, suppose A to be unequal to a and let A = o 

 and B = 



.-.*' = y' 

 .-.jr' + y'^o. 



Now x 1 and y- must each be positive. 

 ..* = oandy = o 

 .-. A = a. and B = i. 



