一道数学题

不查表求值:cos(π/13)+

解: 令cos(π/13)+cos(3π/13)+cos(9π/13)=x,则 x^2=[cos(π/13)]^2+[cos(3π/13)]^2+[cos(9π/13)]^2+2cos(π/13)cos(3π/13)+2cos(3π/13)cos(9π/13)+2cos(9π/13)cos(π/13) =1/2[1+cos(2π/13)]+1/2[1+cos(6π/13)]+1/2[1+cos(18π/13)+cos(2π/13)+cos(4π/13)+cos(6π/13)+cos(12π/13)+cos(8π/13)+cos(10π/13) =3/2-1/2[cos(11π/13)+co...全部

  解: 令cos(π/13)+cos(3π/13)+cos(9π/13)=x,则 x^2=[cos(π/13)]^2+[cos(3π/13)]^2+[cos(9π/13)]^2+2cos(π/13)cos(3π/13)+2cos(3π/13)cos(9π/13)+2cos(9π/13)cos(π/13) =1/2[1+cos(2π/13)]+1/2[1+cos(6π/13)]+1/2[1+cos(18π/13)+cos(2π/13)+cos(4π/13)+cos(6π/13)+cos(12π/13)+cos(8π/13)+cos(10π/13) =3/2-1/2[cos(11π/13)+cos(7π/13)+cos(5π/13)]-[cos(11π/13)+cos(9π/13)+cos(7π/13)+cos(π/13)+cos(5π/13)+cos(3π/13)] 而可另证 cos(11π/13)+cos(7π/13)+cos(5π/13)=1/2-[cos(9π/13)+cos(3π/13)+cos(π/13)], ∴x^2=3/2-1/2*(1/2-x)-1/2 →4x^2-2x-3=0。
   解得,x=(1+根13)/4。(只取正根)。 ∴cos(π/13)+cos(3π/13)+cos(9π/13)=(1+根13)/4。 做得很累啊!。收起