求解高一的三角函数化简题(cosπ/11)(cos2π/11)(cos3π/11)(cos4π/11)(cos5π/11

2个回答

  • cos3π/11

    =cos(π-8π/3)=cos(8π/3)

    sin(16π/11)

    = sin(π+5π/11)

    =-sin(5π/11)

    sin(10π/11)

    = sin(π-π/11)

    =sin(π/11)

    (cosπ/11)(cos2π/11)(cos3π/11)(cos4π/11)(cos5π/11)

    =-(cosπ/11)(cos2π/11)(cos4π/11)(cos8π/11)(cos5π/11)

    =-(sinπ/11)(cosπ/11)(cos2π/11)(cos4π/11)(cos8π/11)(cos5π/11) /(sinπ/11)

    =-(1/2)(sin2π/11)(cos2π/11)(cos4π/11)(cos8π/11)(cos5π/11) /(sinπ/11)

    =-(1/4)(sin4π/11)(cos4π/11)(cos8π/11)(cos5π/11) /(sinπ/11)

    =-(1/8)(sin8π/11)(cos8π/11)(cos5π/11) /(sinπ/11)

    =-(1/16)(sin16π/11)(cos5π/11) /(sinπ/11)

    =(1/16)(sin5π/11)(cos5π/11) /(sinπ/11)

    =(1/32)(sin10π/11)/(sinπ/11)

    =(1/32)(sinπ/11)/(sinπ/11)

    =1/32