Question

高一三角函数化简求值题[(根号3*tan12度)-3]/[sin12度*(4*cos12度平方-2)]

Answer 1

[根号3tan12-3]/{sin12[4(cos12)^2-2]}

=[根号3(sin12/cos12)-3]/{sin12[4(cos12)^2-2]}

=[根号3sin12-3cos12]/{sin12cos12[4(cos12)^2-2]}

=(根号3sin12-3cos12)/{2sin12cos12[2(cos12)^2-1]}

=(根号3sin12-3cos12)/[2cos12sin12*cos24]

=[2根号3sin(12-60)]/[sin24cos24]

=[4根号3sin(12-60)]/sin48

=[-4根号3sin48]/sin48

=-4根号3