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cos(x)+cos(x+d)+……+cos(x+kd)求和的过程
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cos(x)+cos(x+d)+……+cos(x+kd)求和的过程
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答案和解析
对第一个和式, 乘以2sin(d/2), 积化和差: 2cos(x+kd)sin(d/2) = sin(x+(k+1/2)d)-sin(x+(k-1/2)d),
可得2cos(x)sin(d/2)+2cos(x+d)sin(d/2)+...+2cos(x+nd)sin(d/2)
= (sin(x+d/2)-sin(x-d/2))+(sin(x+3d/2)-sin(x+d/2))+...+(sin(x+(n+1/2)d)-sin(x+(n-1/2)d))
= sin(x+(n+1/2)d)-sin(x-d/2)
= 2sin((n+1)d/2)cos(x+nd/2).
于是cos(x)+...+cos(x+nd) = sin((n+1)d/2)cos(x+nd/2)/sin(d/2).
可得2cos(x)sin(d/2)+2cos(x+d)sin(d/2)+...+2cos(x+nd)sin(d/2)
= (sin(x+d/2)-sin(x-d/2))+(sin(x+3d/2)-sin(x+d/2))+...+(sin(x+(n+1/2)d)-sin(x+(n-1/2)d))
= sin(x+(n+1/2)d)-sin(x-d/2)
= 2sin((n+1)d/2)cos(x+nd/2).
于是cos(x)+...+cos(x+nd) = sin((n+1)d/2)cos(x+nd/2)/sin(d/2).
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