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Question Number 206362 by MATHEMATICSAM last updated on 12/Apr/24
sin(π/7) × sin((2π)/7) × sin((3π)/7) = ?
sinπ7×sin2π7×sin3π7=?
Answered by Skabetix last updated on 12/Apr/24
=((√7)/8)
=78
Answered by Berbere last updated on 13/Apr/24
sin((π/7))=sin(((6π)/7));sin(((2π)/7))=sin(((5π)/7));sin(((3π)/7))=sin(((4π)/7))  Z^7 −1=Π_(k=0) ^6 (Z−e^((2ikπ)/7) )  ((Z^7 −1)/(Z−1))=Π_(k=1) ^6 (Z−e^((2ikπ)/7) )=Σz^k   z=1,z−e^((2ikπ)/7) =−e^((ikπ)/7) .2isin(((kπ)/7))  ⇒(−2i)^6 .e^(3iπ) Πsin(((kπ)/7))=6⇒Π_(k=1) ^6 sin(((kπ)/7))=(7/(64))  sin((π/7))sin(((3π)/7))sin(((5π)/7))=(√(7/(64)))=((√7)/8)
sin(π7)=sin(6π7);sin(2π7)=sin(5π7);sin(3π7)=sin(4π7)Z71=6k=0(Ze2ikπ7)Z71Z1=6k=1(Ze2ikπ7)=Σzkz=1,ze2ikπ7=eikπ7.2isin(kπ7)(2i)6.e3iπΠsin(kπ7)=66k=1sin(kπ7)=764sin(π7)sin(3π7)sin(5π7)=764=78

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