（1+cosθ+cos2θ+……＋cosnθ)＋i（sinθ＋sin2θ＋……sinnθ）
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解决时间 2021-02-10 17:45
- 提问者网友：做自己de王妃
- 2021-02-10 08:29
（1+cosθ+cos2θ+……＋cosnθ)＋i（sinθ＋sin2θ＋……sinnθ）
最佳答案
- 五星知识达人网友：煞尾
- 2021-02-10 08:47
原式=1+(cosθ+isinθ)+(cos2θ+isin2θ)+...+(cosnθ+isinnθ)
=1+e^(θi)+e^(2θi)+...+e^(nθi) (欧拉公式)
=[1-(e^(θi))^(n+1)]/(1-e^(θi)
=[1-(e^(θi))^(n+1)]/(1-e^(θi))
=[1-(e^((n+1)θi))]/(1-e^(θi))
=[1-cos(n+1)θ-isin(n+1)θ]/(1-cosθ-isinθ)
=2sin((n+1)θ/2)[sin((n+1)θ/2)-icos((n+1)θ/2)]/[2sin(θ/2)(sin(θ/2)-icos(θ/2))]
=(sin((n+1)θ/2)/(sin(θ/2))[cos((n+1)θ/2-π/2)+isin((n+1)θ/2-π/2)]/(cos(θ/2-π/2)+isin(θ/2-π/2))
=(sin((n+1)θ/2)/(sin(θ/2))(cos(nθ/2)+isin(nθ/2))
=(sin((n+1)θ/2)cos(nθ/2)/(sin(θ/2))+i(sin((n+1)θ/2)sin(nθ/2)/(sin(θ/2))
希望能帮到你！
=1+e^(θi)+e^(2θi)+...+e^(nθi) (欧拉公式)
=[1-(e^(θi))^(n+1)]/(1-e^(θi)
=[1-(e^(θi))^(n+1)]/(1-e^(θi))
=[1-(e^((n+1)θi))]/(1-e^(θi))
=[1-cos(n+1)θ-isin(n+1)θ]/(1-cosθ-isinθ)
=2sin((n+1)θ/2)[sin((n+1)θ/2)-icos((n+1)θ/2)]/[2sin(θ/2)(sin(θ/2)-icos(θ/2))]
=(sin((n+1)θ/2)/(sin(θ/2))[cos((n+1)θ/2-π/2)+isin((n+1)θ/2-π/2)]/(cos(θ/2-π/2)+isin(θ/2-π/2))
=(sin((n+1)θ/2)/(sin(θ/2))(cos(nθ/2)+isin(nθ/2))
=(sin((n+1)θ/2)cos(nθ/2)/(sin(θ/2))+i(sin((n+1)θ/2)sin(nθ/2)/(sin(θ/2))
希望能帮到你！
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- 1楼网友：由着我着迷
- 2021-02-10 09:33
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