1.化简sin²α+cosαcos(π/3+α)-sin²(π/6-α)
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/05/01 16:40:54
1.化简sin²α+cosαcos(π/3+α)-sin²(π/6-α)
2.求值:[sin50°(1+√3tan10°)-cos20°]/cos80°√1-cos20°
第一个根号里面只有3,第二个根号里面是1-cos20°
2.求值:[sin50°(1+√3tan10°)-cos20°]/cos80°√1-cos20°
第一个根号里面只有3,第二个根号里面是1-cos20°
1
sin²α+cosαcos(π/3+α)-sin²(π/6-α)
=sin²α+cosα(cosπ/3cosα-sinπ/3sina)-(sinπ/6cosα-cosπ/6sina)²
=sin²α+cosα(1/2cosα-√3/2sina)-(1/2cosa-√3/2sina)²
=sin²α+(1/2cosα-√3/2sina)(cosa-1/2cosa+3/2sina)
=sin²α+1/4cos²a-3/4son²a=1/4
2
sin50°(1+√3tan10°)-cos20°
=cos10°(sin50°(1+√3tan10°)-cos20°)/cos10°
=(sin50(cos10+√3sin10)-cos20cos10)/cos10
=(2sin50(1/2cos10+√3/2sin10)-cos20cos10)/cos10
=(2sin50cos(60-10)-cos20cos10)/cos10
=(2sin50cos(50)-cos20cos10)/cos10
=(sin100-cos20cos10)/cos10
=(cos10-cos20cos10)/cos10=1-cos20
所以原式=(1-cos20)/cos80°√1-cos20°=√(1-cos20)/cos80°
==√(1-(1-2sin10^2))/cos80°=根号2sin10/cos80=根号2
sin²α+cosαcos(π/3+α)-sin²(π/6-α)
=sin²α+cosα(cosπ/3cosα-sinπ/3sina)-(sinπ/6cosα-cosπ/6sina)²
=sin²α+cosα(1/2cosα-√3/2sina)-(1/2cosa-√3/2sina)²
=sin²α+(1/2cosα-√3/2sina)(cosa-1/2cosa+3/2sina)
=sin²α+1/4cos²a-3/4son²a=1/4
2
sin50°(1+√3tan10°)-cos20°
=cos10°(sin50°(1+√3tan10°)-cos20°)/cos10°
=(sin50(cos10+√3sin10)-cos20cos10)/cos10
=(2sin50(1/2cos10+√3/2sin10)-cos20cos10)/cos10
=(2sin50cos(60-10)-cos20cos10)/cos10
=(2sin50cos(50)-cos20cos10)/cos10
=(sin100-cos20cos10)/cos10
=(cos10-cos20cos10)/cos10=1-cos20
所以原式=(1-cos20)/cos80°√1-cos20°=√(1-cos20)/cos80°
==√(1-(1-2sin10^2))/cos80°=根号2sin10/cos80=根号2
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