Question

求（根号3倍tan12°-3）/[sin12°(4cos�0�512°-2)]的值

Answer 1

（√(3)tan12°-3)/[sin12°(4cos12°^2-2)] =(√(3)sin12°-3cos12°)/[2cos12°sin12°(2cos12°^2-1)]

=2(√(3)[1/2sin12°-√(3)/2 cos12°)]/[cos24°sin24°] =4(√(3)[-sin(60°-12°)/sin48° =-4(√(3)

Answer 2

√3tan12-3
=√3sin12°/cos12-3
=(√3sin12°-3cos12°)/cos12
=2√3sin(12°-60°)/cos12
=-2√3sin48°/cos12°
=-4√3sin24°cos24°/cos12°
=-8√3sin12°cos12°cos24°/cos12°
=-8√3sin12°cos24°

sin12°(4(cos12°)^2-2)
=sin12°(4(cos24° 1)/2-2)
=sin12°*2cos24°
=2sin12°cos24°

所以（√3tan12°-3）/[sin12(4(cos12°)^2-2)]=-4√3