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来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 14:37:50
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(1)sinα=[2tan(α/2)] / [1+(tan α/2)^2] =4/5
cosα=[1-(tan α/2)^2] / [1+(tan α/2)^2] =3/5
sin(β-α)=±√{1 - [cos(β-α)]^2} =±2√6/5
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(1)sinα=[2tan(α/2)] / [1+(tan α/2)^2] =4/5cosα=[1-(tan α/2)^2] / [1+(tan α/2)^2] =3/5sin(β-α)=±√{1 - [cos(β-α)]^2} =±2√6/5∵0<α<π/2<β<π∴sin(β-α)=2√6/5cosβ=cos[(β-α)+α]=cos(β-α)cosα - sin(β-α)sinα = （3-...
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(1)sinα=[2tan(α/2)] / [1+(tan α/2)^2] =4/5cosα=[1-(tan α/2)^2] / [1+(tan α/2)^2] =3/5sin(β-α)=±√{1 - [cos(β-α)]^2} =±2√6/5∵0<α<π/2<β<π∴sin(β-α)=2√6/5cosβ=cos[(β-α)+α]=cos(β-α)cosα - sin(β-α)sinα = （3-8√6）/25∵0<α<π/2<β<π∴β=arccos[（3-8√6）/25]=π-arccos[（-3=8√6）/25](2)用面积公式：S=1/2ab*sinC和余弦定理如下：S=(absinC)/2c^2-(a-b)^2=c^2-a^2-b^2+2ab=2ab(1-cosC)得sinC=4(1-cosC),两边平方后1-(cosC)^2=16(1-cosC)^2(1-cosC)(15+17cosC)=0cosC=-15/17 (cosC=1时C=0,舍去）sinC=8/17由a+b≥2根号(ab)得ab≤1S最大值为S=(absinC)/2≤4/17
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0<α<π/2<β<π，tan(α/2)=1/2,cos(β-α)=2/10 这个sinα= -1/2 β=60