tan12°-3/(4cos²12°-2)sin12°

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

玩的放热发热管如果让工人

楼主题目是不是错了。分子不是tan12°-3而是tan12°-√3

tan12°-3/(4cos²12°-2)sin12°
=tan12°-3/2cos24sin12
=sin12-3cos12/cos24sin24
=2(sin12-3cos12)/sin48
sin48=sin(60-12)=sin60cos12-cos60sin1...