sin a+sin 2a +sin 3a +...+sin na怎么求和?
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sina+sin2a+sin3a+...+sinna怎么求和?sina+sin2a+sin3a+...+sinna怎么求和?sina+sin2a+sin3a+...+sinna怎么求和?采用复数解法：
sin a+sin 2a +sin 3a +...+sin na怎么求和?
sin a+sin 2a +sin 3a +...+sin na怎么求和?
sin a+sin 2a +sin 3a +...+sin na怎么求和?
采用复数解法：设复数z=cos a+cos ai=e^ia,则z^2=e^i2a……z^n=e^ina；对s=z+z^2+z^3+……+z^n求和,利用等比级数求和公式,得s=z(1-z^n)/1-z;按实部虚部展开,因为z^n=cos na+sin nai,则s=(cos a+cos 2a+……+cos na)+(sin a+sin 2a+……sin na)i（1式）,由s=z(1-z^n)/1-z,得s=(cos a+sin ai)(1-cos na-sin nai)/1-cos a-sin ai;计算出s的值,再将实部虚部与1式对应即可.
方法二：原式乘以cos a,得sin acos a+sin 2acos a+……sin nacos a,利用积化和差,得1/2[sin 2a+sin3a+sin a+sin 4a+sin 2a+sin 5a+sin 3a+……+sin(n+1)a+sin(n-1)a]=1/2[sin(n+1)a-sin na-sin a]+原式,即原式*cos a=1/2[sin(n+1)a-sin na-sin a]+原式,则原式＝[sin na+sina-sin(n+1)a]/(2-2cos a)
sin a+sin 2a +sin 3a +...+sin na怎么求和?
sin a+sin 2a +sin 3a +...+sin na怎么求和?
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