(急)求证(cos2π/n)^2+(cos4π/n)^2+.....+(cos2(n-1)π/n)^2+(cos2π)^2=n/2
来源:百度知道 编辑:UC知道 时间:2024/05/17 19:11:01
毕业了,东西都忘的差不多了
cos2π/n)^2 + (cos4π/n)^2+..... + (cos2(n-1)π/n)^2 + (cos2π)^2
=[1+cos(4π/n)]/2 + [1+cos(8π/n)]/2 + ..... + [1+cos(cos4(n-1)π/n)]/2 + [1+cos(cos4nπ/n)]/2
= n/2 + 1/2[cos(4π/n)+cos(8π/n)+...+cos(cos4(n-1)π/n)+cos(cos4nπ/n)]
下面证明:cos(4π/n)+cos(8π/n)+...+cos(cos4(n-1)π/n)+cos(cos4nπ/n ) = 0
欧拉公式:e^ix=cosx+isinx
所以 e^i(4π/n) + e^i(4π/n) + ... + e^i(4(n-1)π/n) + e^i(cos4nπ/n)的实数部分就是
cos(4π/n)+cos(8π/n)+...+cos(cos4(n-1)π/n)+cos(cos4nπ/n )
而:
e^i(4π/n) + e^i(8π/n) + ... + e^i(4(n-1)π/n) + e^i(cos4nπ/n)
= e^i(4π/n)[1-e^i(4nπ/n)]/[1-e^4π]
= 0
所以原式左面= n/2 + 1/2 * 0 = n/2
(急)求证(cos2π/n)^2+(cos4π/n)^2+.....+(cos2(n-1)π/n)^2+(cos2π)^2=n/2
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