174 Of the Division of the Circle 



whence we get 



Cos 4(p = cos-j- 



Sin 4(p = sin --. 



4 



In order to find how many different solutions of the equations 

 (3) are derived from the multiples of b, let all the values of <p and 



- that have different cosines be written in two lines, viz. 



:p:4b, Sb, \2b, ]6b, 20b, 24b, 2Sb, 

 lb, 3b, b, 5b, 6b, 2b, 



i. : b, 2b, 3b, 4b, 5b, bb, 7b. 



From this it is evident that in four instances, viz. when f = 



4b, <p = 8^, 4> = 16^5 and (p = 28b ; cos <p and cos -j , that is, 



X, and y, are unequal : but these four instances give no more 

 than two values of x and two corresponding values of y in the 

 equations (3). In the other three instances, viz. when f=\2b, 



<p = 2Qh, and <p = 24^ ; it appears that cos f = cos --, or x = y; 

 and in these cases the two equations (3) coincide in one, viz. 



x= 8x+— 8a;^ + 1, 

 that is Sx* — Sx^ — x + 1 = 0, 



or, which is the same, 



{4x^+ 2x -1) (2a; + 1) (a; - 1) = 0. 



Now, in the equation 4x'^ + 2x — 1 = 0, the values of x are 

 X = cos 12^ = cos 3b, and x = cos 24b = cos 6b ; and these 

 determine the points of the quindecagon that coincide with the 

 pentagon. In the factor 2a: + I = 0, a = cos 20^ = cos 5b ; and 

 this gives the points of the quindecagon coinciding with the 

 equilateral triangle. The remaining factor a: — 1 = 0, corre- 

 sponds to the cases when ip and — are either both zero, or both 

 multiples of 360^ 



On the whole, setting aside the particular case when x = y, 

 the complete solution of the equations (3) will give six different 

 values of x and as many corresponding values of y. Of these 

 twelve values, two of x and two of y determine those eight 

 points of the quindecagon that coincide neither with the penta- 

 gon nor the equilateral triangle ; and the remaining values of x 

 and y determine all the points of the polygon of seventeen sides. 



In order to solve the equations (3), let x =: -^^r— , y = ' '. 



then, by substitution, 



m — n 



