Question

求值:cos(π/13)+cos(3π/13)+cos(9π/13)

Answer 1

(1+根号13)/4
好像求和符号上下的字母显示的不正确
反正上面都是n
下面都是k=1哦
为方便起见，记cos(π/13)+cos(3π/13)+cos(9π/13)为(1,3,9)13
设(1,3,9)13=x,
(5,7,11)13=y(即cos(5π/13)+cos(7π/13)+cos(11π/13)=y)
利用恒等式
n
∑
cos(2k-1)π/(2n+1)=1/2
K=1
得x+y
=
cos(π/13)+cos(3π/13)+cos(5π/13)
+cos(7π/13)+cos(9π/13)+cos(11π/13)=1/2
{恒等式证明如下:(当然可以略过)
设π/(2n+1)=θ
n
n
sinθ*
∑cos(2k-1)θ
=
∑(sinθ*cos(2k-1)θ)
k=1
k=1
n
=1/2
∑(sin2kθ-sin(2k-2)θ)
K=1
n
n
=1/2
(∑sin2kθ-
∑sin(2k-2)θ)=1/2
sin2nθ
...........(将θ代入)
k=1
k=1
=1/2
sin2nπ/(2n+1)=1/2sin(π-π/(2n+1))
=1/2sinπ/(2n+1)=1/2
sinθ
n
即
∑
cos(2k-1)π/(2n+1)=1/2
}
K=1
同时x*y
=(1,3,9)13
*
(5,7,11)13
=
1/2(6,4,8,6,12,10,2,8,10,4,14,8,4,14,2,16,2,20)
13
=
-1/2(7,9,5,7,1,
3,11,5,
3
,9,
1
,
5,9,1,11,3,11,7)
13
=-1/2(3*(1,3,5,7,9,11)
13)
=-3/4
由x+y=1/2
,xy=-3/4
所以x，y是方程p*p-p/2-3/4=0的两根
所以x=(1+根号13)/4