作业帮z为复数.建立恒等式1+z+z^2+z^3+···+z^n=[1-z^(n+1)]/（1-z）(z不等于1）并导出：1+cosθ+cos2θ+cos3θ+···+cosnθ=1/2+sin(n+1/2)θ/2sin(θ/2)（0
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/28 17:09:34
![z为复数.建立恒等式1+z+z^2+z^3+···+z^n=[1-z^(n+1)]/（1-z）(z不等于1）并导出：1+cosθ+cos2θ+cos3θ+···+cosnθ=1/2+sin(n+1/2)θ/2sin(θ/2)（0](/uploads/image/z/2773763-35-3.jpg?t=z%E4%B8%BA%E5%A4%8D%E6%95%B0.%E5%BB%BA%E7%AB%8B%E6%81%92%E7%AD%89%E5%BC%8F1%2Bz%2Bz%5E2%2Bz%5E3%2B%C2%B7%C2%B7%C2%B7%2Bz%5En%3D%5B1-z%5E%28n%2B1%29%5D%2F%EF%BC%881-z%EF%BC%89%28z%E4%B8%8D%E7%AD%89%E4%BA%8E1%EF%BC%89%E5%B9%B6%E5%AF%BC%E5%87%BA%EF%BC%9A1%2Bcos%CE%B8%2Bcos2%CE%B8%2Bcos3%CE%B8%2B%C2%B7%C2%B7%C2%B7%2Bcosn%CE%B8%3D1%2F2%2Bsin%28n%2B1%2F2%29%CE%B8%2F2sin%28%CE%B8%2F2%29%EF%BC%880)
z为复数.建立恒等式1+z+z^2+z^3+···+z^n=[1-z^(n+1)]/（1-z）(z不等于1）并导出：1+cosθ+cos2θ+cos3θ+···+cosnθ=1/2+sin(n+1/2)θ/2sin(θ/2)（0
z为复数.建立恒等式1+z+z^2+z^3+···+z^n=[1-z^(n+1)]/（1-z）(z不等于1）并导出：1+cosθ+cos2θ+cos3θ+···+cosnθ=1/2+sin(n+1/2)θ/2sin(θ/2)（0
z为复数.建立恒等式1+z+z^2+z^3+···+z^n=[1-z^(n+1)]/（1-z）(z不等于1）并导出：1+cosθ+cos2θ+cos3θ+···+cosnθ=1/2+sin(n+1/2)θ/2sin(θ/2)（0
令z=e^jθ,代入恒等式,取实部就好
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