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求值:(√3tan12°-3)/(4(cos12°)^2-2)sin12°
人气:294 ℃ 时间:2020-05-23 01:29:20
解答
(√3tan12°-3)/[(4(cos12°)^2-2)sin12° ]=√3(tan12°-√3)/[2(cos24°)*sin12° ]
=√3(tan12°-√3)*cos12°/[2(cos24°)*sin12°cos12° ]=2√3(sin12°-√3cos12°)/sin48°
=4√3sin(12°-60°)/sin48°=-4√3
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