你想要当1的时候它就等于0，你想要当0的时候它就等于1，这叫发挥主观能动性可0可1公式

我知道！答案是C

做题蛆滚

给cos的n次方降幂，降幂公式自己网上找

Consider polynomial u7=u6-u5+u4-u3+u2-u for u is the 7th root of -1 form a u cyclic group, rotation for nth won't affect the result. Then consider the real part of the u5+u3+u

https://math.stackexchange.com/questions/140388/how-can-one-prove-cos-pi-7-cos3-pi-7-cos5-pi-7-1-2 有个解答两边sina同乘 改成sina的n次方可破

不知道你学没学过复变函数，转化成指数形式，求等比数列和的实部即可。

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https://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression

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Let x=2cos(pi/7), y=2cos(3pi/7), and z=2cos(5pi/7). Note that x\^n+y\^n+z\^n is a symmetric polynomial in x, y, and z with integer coeffecients. As has been shown in the comments, x+y+z=1. So you only need to show that the other two elementary symmetric polynomials xy+yz+xz and xyz are integers too, then by the The fundamental theorem of symmetric polynomials, x\^n+y\^n+z\^n must be an integer.

To find xy+yz+xz and xyz, note that z=-2cos(2pi/7) and y=-2cos(4pi/7). Then you can express y and z in x and write x+y+z=1 as an equation in x. It turns out to be a quartic equation. You can use the quartic formula to find one simple root and a cubic equation remains. Then write xy+yz+xz and xyz in terms of x as well. Use the quartic equation to reduce them to be cubic in x and their values are given directly by the cubic equation.

夹逼准则试试 集美差不多得了不是你那两片大阴唇😅

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这是啥啊，我看不懂啊

你们数学天天证这逼玩意有什么用啊