sinθ+sin2θ+sin3θ+sin4θ+……+sinnθ=?
答案:5 悬赏:30 手机版
解决时间 2021-11-12 03:10
- 提问者网友：疯子也有疯子的情调
- 2021-11-11 05:53
sinθ+sin2θ+sin3θ+sin4θ+……+sinnθ=?
最佳答案
- 五星知识达人网友：过活
- 2021-11-11 06:25
2sinθsin(θ/2)=cos(θ/2)-cos(3θ/2)
2sin2θsin(θ/2)=cos(3θ/2)-cos(5θ/2)
2sin3θsin(θ/2)=cos(5θ/2)-cos(7θ/2)
2sin4θsin(θ/2)=cos(7θ/2)-cos(9θ/2)
……
2sinnθsin(θ/2)=cos(nθ-θ/2)-cos(nθ+θ/2)
相加得到，
2sin(θ/2)[sinθ+sin2θ+……+sinnθ]=cos(θ/2)-cos(nθ+θ/2)
∴ sinθ+sin2θ+……+sinnθ
=[cos(θ/2)-cos(nθ+θ/2)]/[2sin(θ/2)]
2sin2θsin(θ/2)=cos(3θ/2)-cos(5θ/2)
2sin3θsin(θ/2)=cos(5θ/2)-cos(7θ/2)
2sin4θsin(θ/2)=cos(7θ/2)-cos(9θ/2)
……
2sinnθsin(θ/2)=cos(nθ-θ/2)-cos(nθ+θ/2)
相加得到，
2sin(θ/2)[sinθ+sin2θ+……+sinnθ]=cos(θ/2)-cos(nθ+θ/2)
∴ sinθ+sin2θ+……+sinnθ
=[cos(θ/2)-cos(nθ+θ/2)]/[2sin(θ/2)]
全部回答
- 1楼网友：迷人又混蛋
- 2021-11-11 09:25
等于零肯定错了
- 2楼网友：青灯有味
- 2021-11-11 08:49
最后等于0
- 3楼网友：十年萤火照君眠
- 2021-11-11 07:42
S=sinα+sin2α+sin3α+......+sin nα
2cosα .S =2sinα.cosα+2sin2α.cosα+2sin3α.cosα+...+2sinnα.cosα
=[sin2α + sin0] + [sin3α +sinα]+...+ [sin(n+1)α + sin(n-1)α ]
= sin0 + sin(n+1)α +2S - sinα - sinnα
2(cosα-1)S= sin(n+1)α - sinα - sinnα
S = [sin(n+1)α - sinα - sinnα]/[2(cosα-1)]追问
2cosα .S =2sinα.cosα+2sin2α.cosα+2sin3α.cosα+...+2sinnα.cosα
=[sin2α + sin0] + [sin3α +sinα]+...+ [sin(n+1)α + sin(n-1)α ]
= sin0 + sin(n+1)α +2S - sinα - sinnα
2(cosα-1)S= sin(n+1)α - sinα - sinnα
S = [sin(n+1)α - sinα - sinnα]/[2(cosα-1)]追问
- 4楼网友：夜风逐马
- 2021-11-11 07:17
2sinθsin(θ/2)=cos(θ/2)-cos(3θ/2)
2sin2θsin(θ/2)=cos(3θ/2)-cos(5θ/2)
2sin3θsin(θ/2)=cos(5θ/2)-cos(7θ/2)
2sin4θsin(θ/2)=cos(7θ/2)-cos(9θ/2)
……
2sinnθsin(θ/2)=cos(nθ-θ/2)-cos(nθ+θ/2)
相加得到，
2sin(θ/2)[sinθ+sin2θ+……+sinnθ]=cos(θ/2)-cos(nθ+θ/2)
∴ sinθ+sin2θ+……+sinnθ
=[cos(θ/2)-cos(nθ+θ/2)]/[2sin(θ/2)]追问追答和我的答案一致
2sin2θsin(θ/2)=cos(3θ/2)-cos(5θ/2)
2sin3θsin(θ/2)=cos(5θ/2)-cos(7θ/2)
2sin4θsin(θ/2)=cos(7θ/2)-cos(9θ/2)
……
2sinnθsin(θ/2)=cos(nθ-θ/2)-cos(nθ+θ/2)
相加得到，
2sin(θ/2)[sinθ+sin2θ+……+sinnθ]=cos(θ/2)-cos(nθ+θ/2)
∴ sinθ+sin2θ+……+sinnθ
=[cos(θ/2)-cos(nθ+θ/2)]/[2sin(θ/2)]追问追答和我的答案一致
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