sin的連乘積

sin(π/11)sin(2π/11)sin(3π/11)sin(4π/11)sin(5π/11)的解

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  • 釋塵
    Lv 7
    1 decade ago
    Favorite Answer

    設w = cos(2π/11) + isin(2π/11)

    根據棣美弗定理,則w^n = cos(2nπ/11) + isin(2nπ/11)

    當n = 11時,w^11 = 1

    其中w,w^2,w^3,……………..w^11為x^11 -1 = 0根

    x^11 -1 =(x-1)(1+x+x^2 +………..+x^10) = (x-w)(x-w^2)……………….(x-w^11)

    由此可得1+x+x^2 +………..+x^10 = (x-w)(x-w^2)………………(x-w^10)

    當x = 1時

    11 = (1-w)(1-w^2)……………..(1-w^10)

    則[sin(π/11)sin(2π/11)sin(3π/11)sin(4π/11)sin(5π/11)]^2

    = |(1-w)(1-w^2)………….(1-w^9)(1-w^10)| /1024

    = 11/1024

    所以sin(π/11)sin(2π/11)sin(3π/11)sin(4π/11)sin(5π/11) = √11/32……………(解答)

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