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证明,sina+sin(a+b)+sin(a+2b)+...sin(a+nb)=sin(a+nb/2)sin[(n+1)b/2]/sin(b/2)
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证明,sina+sin(a+b)+sin(a+2b)+...sin(a+nb)=sin(a+nb/2)sin[(n+1)b/2]/sin(b/2)
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答案和解析
[sina+sin(a+b)+sin(a+2b)+..+sin(a+nb)]sin(b/2)
=sinasin(b/2)+sin(a+b)sin(b/2)+sin(a+2b)sinb/2)+...+sin(a+nb)sin(b/2)
=(-1/2)[cos(a+b/2)-cos(a-b/2)+cos(a+3b/2)-cos(a+b/2)+...+cos(a+(2n+1)b/2)-cos(a+(2n-1)b/2)
=(-1/2)[cos(a+(2n+1)b/2)-cos(a-b/2)]
=sin(a+nb/2)sin(n+1)b/2
=sinasin(b/2)+sin(a+b)sin(b/2)+sin(a+2b)sinb/2)+...+sin(a+nb)sin(b/2)
=(-1/2)[cos(a+b/2)-cos(a-b/2)+cos(a+3b/2)-cos(a+b/2)+...+cos(a+(2n+1)b/2)-cos(a+(2n-1)b/2)
=(-1/2)[cos(a+(2n+1)b/2)-cos(a-b/2)]
=sin(a+nb/2)sin(n+1)b/2
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