是在网上找到了关于sin的
sin(π/(2n+1))sin(2π/(2n+1))sin(3π/(2n+1)).sin(nπ/(2n+1))=√(2n +1)/2^n
设Z=cos2π/(2n+1)+ isin2π/(2n+1)
则x^(2n+1)=1的根为1,z,...z^2n
得x^2n+...+x+1=(x-z)(x-z^2)...(x-z^2n)
2n+1=｜(1-z)｜｜(1-z^2)｜...｜(1-z^2n)｜...(1)
又｜(1-z^k)｜=2sinkπ/(2n+1)...(2)
|1-z^k| = |1-(cos(2kπ/(2n+1)) +sin(2kπ/(2n+1)) )|
=|1-cos(2kπ/(2n+1))) -sin(2kπ/(2n+1)) )|
=√((1-2cos(2kπ/(2n+1)) +cos^2 (2kπ/(2n+1))) + sin^2 (2kπ/(2n+1)))
=√(2-2cos(2kπ/(2n+1)) )
=√(4sin^2(kπ/(2n+1))
=2sin(kπ/(2n+1)