DERIVATION OF CUBIC SOLUTION FOR cos(2pi/7) double and triple angle formulas cos(2x) = 2cos2(x) - 1 cos(3x) = (4cos2(x) - 3)cos(x) cos(4x) = 2cos2(2x) - 1 = 8cos4(x) - 8cos2(x) - 1 --------------------------------------------------- sin(3x) = 3sin(x) - 4sin3(x) sin(5x) = sin(3x+2x) = sin(3x)cos(2x) + cos(3x)sin(2x) = (3sin(x) - 4sin3(x))(2cos2(x) - 1) + ((4cos2(x) - 3)cos(x))(2sin(x)cos(x)) = 6sin(x)cos2(x) -8sin3(x)cos2(x) -3sin(x) + 4sin3(x) +8sin(x)cos4(x) -6sin(x)cos2(x) = -8sin3(x)cos2(x) + 4sin3(x) + 6sin(x)cos2(x) -3sin(x) -6sin(x)cos2(x) +8sin(x)cos4(x) = 4sin3(x) (1- 2cos2(x)) + sin(x)( -3 +8cos4(x)) = 4sin3(x) (2sin2(x)-1) + sin(x)( -3 +8(1-sin2(x))^2) = 4sin3(x) (2sin2(x)-1) + sin(x)( 8(1-2sin2(x)+sin4(x)) -3 ) = 8sin5(x) -4sin3(x) + 8sin5(x) -16sin3(x) + 8sin(x) -3sin(x) = 16sin5(x) -20sin3(x) + 5sin(x) check sin(5x) = 5sin(x) - 20sin2(x) + 15sin5(x) check cos(5x) = 16cos4(x) - 20cos3(x) + 5cos(x) //mathforum.org/dr.math/faq/formulas/faq.trig.html 5cos(5x) = 80sin4(x)cos(x) -60sin2(x)cos(x) +5cos(x) cos(5x) = (15 sin4(x) - 12 sin2(x) + 1) cos(x) = (15 (1- cos2(x))^2 -12(1-cos2(x)) + 1) cos(x) ... not finished yet, but not needed --------------------------------------------------- cos(6x) = 32cos6(x) -48cos4(x) + 18cos2(x) - 1 check //mathforum.org/dr.math/faq/formulas/faq.trig.html sin(6x) = 2sin(3x)cos(3x) = 2(3sin(x) - 4sin3(x)) (4cos2(x) - 3) cos(x) = 2sin(x)cos(x)(-16sin2(x)cos2(x) + 12(cos2(x)+sin2(x)) -9) 2sin(x)cos(x) (16sin4(x) -16sin2(x) + 3) --------------------------------------------------- cos(2pi/7) - cos( 5pi/7) = 0 cos(2pi/7) - cos(12pi/7) = 0 !! [sept.txt 304] cos(3pi/7) - cos(4pi/7) = 0 cos(pi/7) - cos(6pi/7) = 0 [in sept.t] - cos(pi/7) = - cos(6pi/7) = -3cos(2pi/7) + 4cos3(2pi/7) = -6cos2(pi/7) + 3 + 4(2cos2(pi/7)-1)^3 = 32cos6(pi/7) - 48cos4(pi/7) + (24-6)cos2(pi/7) - 4 + 3 32cos6(pi/7) -48cos4(pi/7) + 18cos2(pi/7) + cos(pi/7) - 1 = 0 - 1 1 [1/2+cos(pi/7)/2] 9 2cos2(pi/7) = cos(2pi/7) + 1 -24 2cos4(pi/7) = cos2(2pi/7) + 2cos(2pi/7) + 1 16 2cos6(pi/7) = cos3(2pi/7) + 3cos2(2pi/7) + 3cos(2pi/7) + 1 16 3 9 1 0 16cos3(2pi/7) + 3cos2(2pi/7) + 9cos(2pi/7) + cos(pi/7) = 0 we need all odd or even terms of the same angle we can get 2pi/7 and 12pi/7 by going to 5pi/7 and then going around again to 12pi/7. [sept.txt 290-327] apply double and triple angle formulas to 6pi/7 pi was 3.14159 cos(2pi/7) = cos(12pi/7) =0.623489802 = (4cos2(4pi/7) -3) cos(4pi/7) = (4(2cos2(2pi/7)-1)^2 -3) (2cos2(2pi/7)-1) y = (4(2y^2-1)^2-3)(2y^2-1) = 4(2y^2-1)^3 - 3(2y^2-1) y = 4(8y^6 - 12y^4 + 6y^2 -1) - 6y^2 +3 32y^6 -48y^4 +18y^2 -y -1 = 0 (y - 0.623489802) (A) x^6 /2 - 3 x^4 + (9/2) x^2 - x /2 -1 = 0 (x-1.246979603) x^6 - 6 x^4 + 9 x^2 -x -2 = 0 !!! (x^3 - 3x)(x^3 -3x) - (x+2) =0 x^2(x^2-3)^2 - (x+2) =0 if P(x) = ax^n + ... + z and x=p/q is a rational zero of P(x) then p divides z and q divides a http://www.sosmath.com/algebra/factor/fac10/fac10.html !! x(32x^5 -48x^3 +18x -1) = 1 p() = 1q^5 p divides 2 +-2 a0 32p^5 = q() q divides 1 +-q is +-1 an +- p/q = 2,1 64-96+36 -+2 -2 = 0 x-2=0 (x-2)(x^5 +ax^4 +bx^3 +cx^2 +dx +1) =0 a-2=0 b-2a=-6 c-2b=0 d-2c=9 1-2d=-1 x^5 +2x^4 -2x^3 -4x^2 +dx +1 =0 (x^2 +ax -1) * ( x^3 +x^2 -2x -1) =0 1+a=2 -2+a=-2 -1-2a-1=-4 -a+2=1 (x^2 +x -1) * ( x^3 +x^2 -2x -1) =0 !! [sept.txt 290-327] www.sosmath.com/algebra/factor/fac10/fac10.html x^3 + x^2 - 2x - 1 = 0 wiki/Heptagon - - - - - - - - - - - - - - - - - - - - - - 336 [sept.txt 366-454] x =2cos(2pi/7) x^3 + x^2 - 2x - 1 = 0 in wiki/Heptagon (B) checked numerically // 8 4 -4 -1 >< 0 (x-1.2470)=0 x = y - 1/3 (y-1.5803)=0 y^3 -y^2 +y/3 +1/27 +y^2 -2/3 y + 1/9 -2y +2/3 -1 =0 y^3 - 1/27 + 1/3 y - 8/3 y + 2/9 = 0 y^3 - 7/3 y - 7/27 = 0 y^3 - 7/3 y = 7/27 5.29167 4.066067 0.259259259 0.96634 y^3 - 3ab y = a^3 - b^3 1.5803 numerically checks extrema stationary points 3y^2 - 7/3 = 0 y = +- [7]/3 x = ([7]-1)/3,-([7]+1)/3 y=a-b vector y^3=a^3-3a^2b+3ab^2-b^3 y^3=a^3+3ab(a-b)-b^3 y^3-3aby=a^3-b^3 y^3-3my = n (x+1/3)^3-3m(x+1/3)-n =0 x^3 +x^2 +1/3 x + 1/27 -3mx -m -n =0 x^3 +x^2 +(1/3-3m) x +1/27 -m -n = 0 i.e. n +m = -d +1/27 1/3 - 3m = c m = 1/9 - c/3 e.g. m = 1/9 +7/9 = 8/9 n = 1/27 -8/9 +7/27 = 18/9 constraint: 3&2 coefficints equal remedy: scale so that x=sy x^3 + (a^(1/2)x)^2 + ... s^3y^3 + as^2y^2 + ... s = a a^3 - b^3 = 7/27 =n 3ab = 7/3 =3m m=?7/9 b=?/3a m^3/n^2 = (7/9)^3/(7/27)^2 = 7 a^6 - n a^3 + m^3 = 0 a^6 - 7/27 a^3 - (7/9)^3 = 0 a^3 - n a^3 + m =0 a^3 = (n +- [n^2 - 4m^3])/2 a = n^(1/3) (1 +- [1 - 4 m^3/n^2])^(1/3) /2^(1/3) b^3 = (n +- [n^2 - 4m^3])/2 -n = (-n +- [n^2 - 4m^3])/2 b = n^(1/3) (-1 +- [1 - 4 m^3/n^2])^(1/3) /2^(1/3) y = a-b = n^(1/3)/2^(1/3) (( 1 +- [1 - 4 m^3/n^2])^(1/3) -(-1 +- [1 - 4 m^3/n^2])^(1/3)) x=y-1/3= n^(1/3)/2^(1/3) (( 1 +- [1 - 4 m^3/n^2])^(1/3) -(-1 +- [1 - 4 m^3/n^2])^(1/3)) -1/3 if(4 m^3/n^2 < 1) printf("error 374"); (7/9)^3/(7/27)^2 = 7 y = (7/27)^3/2 ((1+-[1-4*7])^(1/3)-(-1+-[1-4*7])^(1/3)) x = (7/27)^3/2 ((1+-[1-4*7])^(1/3)-(-1+-[1-4*7])^(1/3)) -1/3 a = (n/2)^(1/3) ( 1 +- [ 1 - 4*m^3/n^2 ])^(1/3) b = (n/2)^(1/3) (-1 +- [ 1 - 4*m^3/n^2 ])^(1/3) r=[1+[4m^3/n^2-1]] thb - tha = pi - 2*atan(4m^3/n^2-1) (nr)^(1/3)2^(5/3)sin((thb-tha)/6) 4(7/9)^3/(27/7)^2 -1 = 27 r =[1+[4*7-1]] = [1+27], atan([3^3]) y = (7/27 [1+27] /2)^(1/3) 2*sin((pi-2*atan([27])/3)/2) [sept.t 336-454] ------------------------------------------------------------------ 2cos(2pi/7) = (7/27 [1+3^3] /2)^(1/3) 2*sin((pi-2*atan([3^3])/3) - 1/3 (c) 2009 Wm.C.Corwin