求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11

1个回答

  • cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11

    =(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

    解题关键在于利用与构造sin2x=2SinxCosx.如果不明白,等你追问.