这道三角函数变式题怎么做
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sin40°(tan10°-√3)
=sin40°(sin10°/cos10° -√3)
=sin40°(sin10°-√3cos10°)/cos10°
=2sin40°[(1/2)sin10°-(√3/2)cos10°)]/cos10°
=2sin40°[cos60°sin10°-sin60°cos10°)]/cos10°
=2sin40°sin(-50°)/cos10°
=-2sin40°cos40°/cos10°
=-sin80°/cos10°
=-cos10°/cos10°
=-1
=sin40°(sin10°/cos10° -√3)
=sin40°(sin10°-√3cos10°)/cos10°
=2sin40°[(1/2)sin10°-(√3/2)cos10°)]/cos10°
=2sin40°[cos60°sin10°-sin60°cos10°)]/cos10°
=2sin40°sin(-50°)/cos10°
=-2sin40°cos40°/cos10°
=-sin80°/cos10°
=-cos10°/cos10°
=-1
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回答的不是变式把
哦不好意思
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