T R I G O N O M E T R I C   F U N C T I O N S   A S   R O O T S   O F   P O L Y N O M I A L S
Mathematical Curiosities Index	Homepage

For the function:- cos (2pk) m let (2pk) = q in radians m Unless k = m or a multiple of m, when cos q = 1, or k = m or a multiple of m, when cos q = -1 :- 2 2 Then if m is an odd integer, and k is an integer, let m-1 = n 2 cos q is then a root of the polynomial:- 2n-4(n-3)(n-2)xn-4 2nxn + 2n-1xn-1 - 2n-2(n-1)xn-2 - 2n-3(n-2)xn-3 + ————————————————— 2! 2n-5(n-4)(n-3)xn-5 2n-6(n-5)(n-4)(n-3)xn-6 + ————————————————— - —————————————————————— 2! 3! 2n-7(n-6)(n-5)(n-4)xn-7 2n-8(n-7)(n-6)(n-5)(n-4)xn-8 - —————————————————————— + ——————————————————————————— 3! 4! 2n-9(n-8)(n-7)(n-6)(n-5)xn-9 2n-10(n-9)(n-8)(n-7)(n-6)(n-5)xn-10 + ——————————————————————————— - —————————————————————————————————— 4! 5! 2n-11(n-10)(n-9)(n-8)(n-7)(n-6)xn-11 - —————————————————————————————————— + ... ± 1 = 0 5! (discarding terms when the exponent becomes negative). Or if m is an even integer, and k is an integer, let m-2 = n 2 then cos q is a root of the polynomial :- 2n-4(n-3)(n-2)xn-4 2n-6(n-5)(n-4)(n-3)xn-6 2nxn - 2n-2xn-2 + ————————————————— - —————————————————————— 2! 3! 2n-8(n-7)(n-6)(n-5)(n-4)xn-8 + ——————————————————————————— - ... = 0 4!     This is not a proof. In fact calculations to 16 or so significant figures are probably insufficient when m becomes larger than about 50.     If you write down the coefficients of this series of equations in the form of a triangle you can see a connection with the ‘Fibonacci’ series.

Homepage

Mathematical Curiosities Index