282 DEMONSTRATION OF THE COTESIAN THEOREM. 



Demonstration x » {m X + 3 c' 



of Cotes' 

 theorem. 



Demonstration ^ a. „ / »t .r -f 3 c \ 



of Cotes's H~ 2 <? y cos. (^ J + 1 



&c. 



<i .v —zqy co s - ^— — ■— J + i 



Since (lien our first formula of the 2 mth degree has for 

 its divisors the above m formulte of the 2d degree; it fol- 

 lows, from the nature of equations, that it is equal to the 

 product of these m formula?. 



Now by making m x — 0, aud 771 x — — » we have 2 cos. 

 mxzz -4- 1, and the general form is reduced to this parti- 

 cular one, 



(q m y m ± 1) 4 , 



which is the Cotesian theorem. 



For in the first case, making mx — 0, it becomes 

 (ff-l)*, 

 having for its divisors, 



£tf~ 2 4y cos. QfifyA* * 



g*y* — 2<7ycos.(^) + 1 



?V~2?ycos.(i~)+l 

 &c. &c. 



And secondly, making mx — — , we have for our formula 



(/i m y m + 1)% 

 its divisors being 



** y»- 2 9 y cos. (£-)+! 



&c. &c. 



Or 



