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  • RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11

    RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11
    數(shù)學(xué)人氣:622 ℃時(shí)間:2020-03-20 16:12:14
    優(yōu)質(zhì)解答
    原式
    =(cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11*2sinπ/11)/2sinπ/11
    =(sin2π/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
    =(1/2*sin4π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
    =(1/4*sin8π/11*cos3π/11*cos5π/11)/2sinπ/11
    =(1/4*sin3π/11*cos3π/11*cos5π/11)/2sinπ/11
    =(1/8*sin6π/11*cos5π/11)/2sinπ/11
    =(1/8*sin5π/11*cos5π/11)/2sinπ/11
    =(1/16*sin10π/11)/2sinπ/11
    =1/32
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