Question

化简sin40度（tan10度减根号3）跪求

Answer 1

解：
sin40°（tan10°-√3）
=sin40°(sin10°/cos10°-√3)
=sin40°(sin10°-√3cos10°)/cos10°
=2sin40°[sin10°cos60°-cos10°sin60°]/cos10°
=2sin40*sin(-50°)/cos10°
=-2sin40°cos40°/cos10°
=-sin80°/cos10°
=-1

Answer 2

解：
sin40°（tan10°-√3）
=sin40°*2*(1/2*sin10°-√3/2*cos10°)/cos30°
=2sin40°(cos60°sin10°-sin60°cos10°)/cos10°
=2sin40°[sin10°cos60°-cos10°sin60°]/cos10°
=2sin40*sin(-50°)/cos10°
=-2sin40*sin50°/cos10°
=-2sin40°cos40°/cos10°
=-sin80°/cos10°
=-cos10°/cos10°
=-1

Answer 3

我认为以下俩回答都是正确的！！！