tan12°-3/(4cos²12°-2)sin12°
3个回答

做了半天,题目应该错了,是(tan12°-√3)/(4cos²12°-2)sin12°

tan12°-√3/(4cos²12°-2)sin12°

=(tan12°-√3)/[2(cos²12-1)*sin12°]

=(tan12°-√3)/(2sin12°cos24°)

=(sin12°-√3cos12°)/(2sin12°cos12°cos24°)

=2[sin12°*(1/2)-cos12°*(√3/2)]/(sin24°cos24°)

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

=4sin(12°-60°)/sin48°

=-4sin48°/sin48°

=-4