正弦連乘積公式
證明:
cosθ+isinθ=eiθ⇒sinθ=eiθ−e−iθ2i⇒sinkπn=eikπn−e−ikπn2i=e−ikπn2i(ei2kπn−1)⇒sinπn⋅sin2πn⋯sin(n−1)πn=e−iπn(1+2+⋯+(n−1))(2i)n−1(ei2πn−1)(ei4πn−1)⋯(ei2(n−1)πn−1)=e−iπ(n−1)2(2i)n−1(ei2πn−1)(ei4πn−1)⋯(ei2(n−1)πn−1)=(2i)1−ne−iπ(n−1)2(ω−1)(ω2−1)⋯(ωn−1−1),其中ω=ei2πn=21−n⋅eiπ2(1−n)⋅e−iπ(n−1)2(ω−1)(ω2−1)⋯(ωn−1−1)(∵i=eiπ2⇒i1−n=eiπ2(1−n))=21−nei(1−n)π(ω−1)(ω2−1)⋯(ωn−1−1)=21−n(−1)(1−n)(ω−1)(ω2−1)⋯(ωn−1−1)=(−2)1−n(ω−1)(ω2−1)⋯(ωn−1−1)=(−2)1−n(−1)1−n(1−ω)(1−ω2)⋯(1−ωn−1)=21−n(1−ω)(1−ω2)⋯(1−ωn−1)令f(x)=xn−1=(x−1)(1+x+x2+⋯+xn−1),則f(x)=0的根為1,ω,ω2,…,ωn−1,其中ω=ei2πn⇒f(x)=(x−1)(1+x+x2+⋯+xn−1)=(x−1)(x−ω)(x−ω2)⋯(x−ωn−1)⇒1+x+x2+⋯+xn−1=(x−ω)(x−ω2)⋯(x−ωn−1)x=1代入上式⇒n=(1−ω)(1−ω2)⋯(1−ωn−1)⇒sinπn⋅sin2πn⋯sin(n−1)πn=21−n(1−ω)(1−ω2)⋯(1−ωn−1)=21−n⋅n=n2n−1,故得證
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