sin a+sin 2a +sin 3a +...+sin na怎么求和?
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sin a+sin 2a +sin 3a +...+sin na怎么求和?
/>利用积化和差公式,达到裂项的效果.
2sinka*sin(a/2)
=-cos[(k+1/2)a]+[cos(k-1/2)a]
∴ 2sin(a/2)*(sin a+sin 2a +sin 3a +...+sin na)
= 2sina*sin(a/2)+2sin2a*sin(a/2)+2sin3a*sin(a/2)+.+2sin(na) *sin(a/2)
=[cos(a/2)-cos(3a/2)]+[cos(3a/2)-cos(5a/2)]+[cos(5a/2)-cos(7a/2)]+.+[cos(na-a/2)-cos(na+a/2)]
=cos(a/2)-cos(na+a/2)
∴ sin a+sin 2a +sin 3a +...+sin na=[cos(a/2)-cos(na+a/2)]/[2sin(a/2)]
2sinka*sin(a/2)
=-cos[(k+1/2)a]+[cos(k-1/2)a]
∴ 2sin(a/2)*(sin a+sin 2a +sin 3a +...+sin na)
= 2sina*sin(a/2)+2sin2a*sin(a/2)+2sin3a*sin(a/2)+.+2sin(na) *sin(a/2)
=[cos(a/2)-cos(3a/2)]+[cos(3a/2)-cos(5a/2)]+[cos(5a/2)-cos(7a/2)]+.+[cos(na-a/2)-cos(na+a/2)]
=cos(a/2)-cos(na+a/2)
∴ sin a+sin 2a +sin 3a +...+sin na=[cos(a/2)-cos(na+a/2)]/[2sin(a/2)]
sin a+sin 2a +sin 3a +...+sin na怎么求和?
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