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
