Question

sin a+sin 2a +sin 3a +...+sin na怎么求和?

Answer 1

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)]