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sin40°(tan10°-√3)=?
人气:304 ℃ 时间:2020-05-21 12:30:59
解答
tan10-√3
=sin10/cos10-√3
=(sin10-√3cos10)/cos10
=2sin(10-60)/cos10
=-2sin50/cos10
sin40(tan10-√3)
=-(2sin40sin50)/cos10
=-[cos(50-40)-cos(50+40)]/cos10
=-cos10/cos10
=-1
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