Question

sinθ+sin2θ+sin3θ+sin4θ+……+sinnθ=?

Answer 1

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)]

Answer 2

最后等于0

Answer 3

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

Answer 4

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

Answer 5

等于零肯定错了