tan12°-3)/sin12°*[4(cos12°)^2-2]

1个回答

  • 因为 4(cos12°)^2-2]=2[2(cos12°)^2-1] =2cos24°

    (tan12°-根号3)=sin12°/cos12°-根号3=(sin12°-根号3cos12°)/cos12°

    所以(tan12°-根号3)/sin12°*[4(cos12°)^2-2]

    =(sin12°-根号3cos12°)/2sin12°cos12°*cos24°

    =(sin12°-根号3cos12°)/sin24°*cos24°

    =2(sin12°-根号3cos12°)/2sin24°*cos24°

    =2(sin12°-根号3cos12°)/sin48°

    =4(1/2*sin12°-根号3/2*cos12°)/sin48°

    =4(cos60°*sin12°-sin60°*cos12°)/sin48°

    =-4(sin60°*cos12°-cos60°*sin12°)/sin48°

    =-4sin(60°-12°)/sin48°

    =-4sin48°/sin48°

    =-4.