cos2π/2n+1+cos4π/2n+1+......+cos2nπ/2n+1=
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解决时间 2021-03-26 07:55
- 提问者网友：凉末
- 2021-03-25 11:40
cos2π/2n+1+cos4π/2n+1+......+cos2nπ/2n+1=
最佳答案
- 五星知识达人网友：爱难随人意
- 2021-03-25 13:09
将所求式乘以2sin(π/(2n+1))得:
2sin(π/(2n+1))cos(2π/(2n+1))+2sin(π/(2n+1))cos(4π/(2n+1))+...+2sin(π/(2n+1))cos(2nπ/(2n+1))
= (sin(3π/(2n+1))-sin(π/(2n+1)))+(sin(5π/(2n+1))-sin(3π/(2n+1)))+...
+(sin((2n+1)π/(2n+1))-sin((2n-1)π/(2n+1)))
= sin((2n+1)π/(2n+1))-sin(π/(2n+1))
= sin(π)-sin(π/(2n+1))
= -sin(π/(2n+1)).
由sin(π/(2n+1)) ≠ 0, 两边可除以2sin(π/(2n+1))得:
cos(2π/(2n+1))+cos(4π/(2n+1))+...+cos(2nπ/(2n+1)) = -1/2.
2sin(π/(2n+1))cos(2π/(2n+1))+2sin(π/(2n+1))cos(4π/(2n+1))+...+2sin(π/(2n+1))cos(2nπ/(2n+1))
= (sin(3π/(2n+1))-sin(π/(2n+1)))+(sin(5π/(2n+1))-sin(3π/(2n+1)))+...
+(sin((2n+1)π/(2n+1))-sin((2n-1)π/(2n+1)))
= sin((2n+1)π/(2n+1))-sin(π/(2n+1))
= sin(π)-sin(π/(2n+1))
= -sin(π/(2n+1)).
由sin(π/(2n+1)) ≠ 0, 两边可除以2sin(π/(2n+1))得:
cos(2π/(2n+1))+cos(4π/(2n+1))+...+cos(2nπ/(2n+1)) = -1/2.
全部回答
- 1楼网友：長槍戰八方
- 2021-03-25 14:19
-1/2，可以用复数根对应关系/
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