Question

三角函数问题

设△ABC的三个角满足9A=3B=C
求cosA+cosB+cosC
这是一道高中奥数题。我知道一种很麻烦的解法，觉得不是很清晰、美观。
有以下两种思路，但做到一半就不会了，求高手指教。
1.利用公式cosA+cosB+cosC=1+4sin½A sin½B sin½C求解
2.作复数z1，z2，z3为△ABC的三个顶点，寻找方程z³+c1 z²+c2 z+c3=0
使z1，z2，z3为它的三个根，于是c1的实部即为所求式子的值。

Answer 1

法1.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.
法2
cos(π/13)+cos(3π/13)+cos(9π/13)=x
cos(5π/13)+cos(7π/13)+cos(11π/13)=y
x+y=1/2
x*y=（cos(π/13)+cos(3π/13)+cos(9π/13)）（cos(5π/13)+cos(7π/13)+cos(11π/13)）
=-3/2（cos(π/13)-cos(2π/13)+cos(3π/13)-cos(4π/13)+cos(5π/13)-cos(6π/13)）
=-1/2（cos(π/13)+cos(3π/13)+cos(9π/13)+cos(5π/13)+cos(7π/13)+cos(11π/13)）=-3/4
x=(1+根13)/4

这里用到恒等式
证明：cos(2π/2n+1)+cos(4π/2n+1)+……+cos(2nπ/2n+1)=1/2至于证明估计您会

Answer 2

因为A=C/9,B=C/3,
所以A=pi/13,b=3*pi/13,C=9*pi/13
cosA+cosB+cosC=cos(pi/13)+cos(3*pi/13)+cos(9*pi/13)

追问

小学生也能算到这一步。

Answer 3

A=3 B=9 C

Answer 4

√3/2

追问

你以为这是没什么水平的高考题？猜个数就八九不离十了？