作业帮令n趋近于无穷大,且n存在,求sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=?.
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令n趋近于无穷大,且n存在,求sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=?.
令n趋近于无穷大,且n存在,求sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=?.
令n趋近于无穷大,且n存在,求sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=?.
是下面这个题吧：
Lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]
观察：可以看出,实际上就是将区间[0,1]分成n等分,对函数y=sinπx.在每个区间点上求面积,然后求和.
很明显,由定积分的定义可知：
这和定积分∫sinπxdx x从0到1是等价的
所以
Lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=∫sinπxdx
=-1/πcosπx|0,1
=2/π
设sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=A
对上式两边同时乘2sin[π/(2n)]，则
左边第一项：
2sin[π/(2n)]·sin(π/n)=cos[π/n-π/(2n)]-cos[π/n+π/(2n)]=cos[π/(2n)]-cos[3π/(2n)]
左边第二项：
2sin[π/(2n)]·sin...
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设sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=A
对上式两边同时乘2sin[π/(2n)]，则
左边第一项：
2sin[π/(2n)]·sin(π/n)=cos[π/n-π/(2n)]-cos[π/n+π/(2n)]=cos[π/(2n)]-cos[3π/(2n)]
左边第二项：
2sin[π/(2n)]·sin(2π/n)=cos[2π/n-π/(2n)]-cos[2π/n+π/(2n)]=cos[3π/(2n)]-cos[5π/(2n)]
左边第三项：
2sin[π/(2n)]·sin(3π/n)=cos[3π/n-π/(2n)]-cos[3π/n +π/(2n)]=cos[5π/(2n)]-cos[7π/(2n)]
……
左边第n项：
2sin[π/(2n)]·sin[(n)π/n]=cos[(nπ/n-π/(2n)]-cos[nπ/n+π/(2n)]=cos[(2n-1)π(2n)]- cos[(2n+1)π/(2n)]
故：2sin[π/(2n)]·[sin(π/n)+sin(2π/n)+......+sin(nπ/n)]
=cos[π/(2n)]-cos[(2n+1)π/(2n)]
=cos[π/(2n)]-cos[π+π/(2n)]
=2cos[π/(2n)]
故：2sin[π/(2n)][sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=2sin[π/(2n)]A
2cos[π/(2n)]=2sin[π/(2n)]A
A=cos[π/(2n)]/{sin[π/(2n)]}
=cot[π/(2n)]
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