计算sin12度(4cos12度的平方-2)分之tan12度-根号3

√3tan12-3
=√3sin12/cos12-3
=(√3sin12-3cos12)/cos12
=2√3sin(12-60)/cos12
=-2√3sin48/cos12
=-4√3sin24cos24/cos12
=-8√3sin12cos12cos24/cos12
=-8√3sin12cos24
sin12(4(cos12)^2-2)
=sin12(4(cos24+1)/2-2)
=sin12*2cos24
=2sin12cos24
所以√3tan12-3/[sin12(4(cos12)^2-2)]=-4√3