soal matematika simple tapi edan

Akar-akar persamaan ax^3 + bx^2 + cx + d = 0 adalah cos(2Pi/7), cos(4Pi/7) dan cos(6Pi/7), dimana a, b, c dan d adalah bilangan bulat. Tentukan nilai a, b, c dan d.


solusi


 misal x = 2pi/7
2sinx.cosx = sin2x
2sinx.cos3x = sin 4x - sin2x
2sinx.cos5x = sin6x - sin4x
----------------------------------(+)
2sinx(cosx+cos3x+cos5x)=sin6x=-sinx
cosx+cos3x+cos5x=-1/2

cos2pi/7 + cos6pi/7 + cos10pi/7 = -1/2

utk cos10pi/7 = cos(pi + 3pi/7) = -cos3pi/7 = -(cos(pi-4pi/7) = -(-cos4pi/7) = cos4pi/7

shg cos2pi/7 + cos4pi/7 + cos6pi/7 = -1/2


x = 2pi/7
cosx.cos3x.cos5x = 1/2(cos4x+cos2x).cos5x
=1/2(cos4x.cos5x + cos2x.cos5x)
=1/4(cos9x + cosx + cos7x + cos3x)
=1/4(cos2x + cosx + 1 + cos3x)
=1/4(1 + cosx + cos2x + cos3x)
=1/4(1 + cos2pi/7 + cos4pi/7 + cos6pi/7)
= 1/4(1 + (-1/2))
=1/8


cosx.cos3x + cosx.cos5x + cos3x.cos5x =
1/2[cos4x + cos2x + cos6x + cos4x + cos 8x + cos 2x]=
1/2[2.cos2x + 2.cos4x + cos6x + cos8x]
karena cos8x = cos(7x+x)=cosx=cos(7x - x)=cos6x, maka didapat
=1/2[2.cos2x + 2.cos4x + 2cos6x]
=cos2x + cos4x + cos6x
=cos(7x-2x) + cos(7x-4x) + cos(7x - 6x)
=cos5x + cos3x + cosx = -1/2


diperoleh
-b/a = -1/2
c/a = -1/2
-d/a = 1/8

sehingga polinom tsb
x^3 + 1/2 x^2 - 1/2 x + 1/8=0 atau 8x^3 + 4x^2 - 4x + 1=0