Question

（�tan12°-3）/sin12°*[4(cos12°)^2-2]

Answer 1

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。