t , ; , MATHEMATICS. TRIGONOMETRY. [EXPONENTIAL EQUATIONS. 



/ T .in W <J~3+ / * 



_i_ ^ l sin. %H/ - 



ln . - cot. 30 







'- 



IT + /H1 sin. ----cos. 210 + J ln 

 6 fl 



co. W^ sin. 

 6 



270'+ ./-I sin. 270 --- '- 1 



~~ 



il + V^lsin. - - cos. 330 + J=lu*. 330" - 



Similarly if we have to find the roots of the Equation z- 1 =0. Since cos^2j>*-+ y 1 sin. 2pir = 1 



we shall have all the roots given in the formula x = cos. -?-{. J 1 sin. 



n n 



(56). To resolve z 1 " - ' into /actors. * 



If . a* 1 are the roots of the Equation z-" 1 = 0. Then z*" 1 = (z <x ) (z ,) 



(x-a 2 )....(x-a*). 



Now all the roots of the Equation are included in the expression cos. ~^+ >J 1 -g ' 



If then we suppose p n. and a f be the corresponding root a p = cos. ^ + J 1 sin. ~-- 



Now observe that there will be another root a*"-* which will equal cos 





lain. - 



x cos. ?JL_^3Tl sin. ^-J f x cos. ^+ V^lsin.^ ) 



Hence a*"-' = cos.?^ J 1 sin.^~ where p ^ n, and so we have a pair of factors z a? and x a* 



w 



which are equal to ( 



= x 3 2x cos. ^ + 1, by multiplication. 

 n 



Where p may take any value, 1-2-3 n 1. It will be observed that | B __^>J.l, ' ( x ")(* "^ = a;t * 



and .'. z** 1 = (z 1 1) ( x* 2x cos. + 1 ) ( X s 2x cos. + 1 J . . . I z* 2x cos. + 1 J (52) 



CoroL Hence, z*" a*" = (z 3 a 2 ) ( z 3 2azcos. - + a> ) +(* 2azcos. - |-o 1 J...(53) 



/ \ f 2ir \ /, n 1 ir . , \ 



For since v 2 "" 1 = (v 2 1)1 I/ 2 2ucos. (- 1 1 y 3 2ycos 1-1 1 .... I y" 2ycos. 



\ J\ n J \ I 



Now as there are n factors, if we clear the equation of fractions, multiplying the right-hand side by o 4 " will bo the 

 same thing as multiplying each factor by o*. 



Hence : . _ . 



aJ. a*" =(x 1 a' ) ( x' 2axcos. ^+a> )...( x' 2ox cos. l^^ll? + a ] 



V / V / \ n / 



By a precisely similar process we shall be able to prove that 



And hence 



'-2 ax 



a' ..... (55). 



* If a,, a lf a, .... a. are the roots of the equation, 

 Then 



- (x a,) (x a.) (x <i s ) (x o,,) for all 



values of x. 



For if we actually divide x"~i by x a,, we shall 

 obtain a remainder a,"" 1 ; now, since a, is a root of the 



equation x"~' =0, we have a," 1 =0, . e., x" t is 

 divisible byx a,, or x o,, isa factor of x"~>. Simi- 

 larly x a 2 . x a, . . . x On are each factors of x"-', 

 and since x"~ l cannot have more than n factors, it is 

 plain that 



x-i =(x Ol )(x d 2 ) (x a 



