1.化简sin²α+cosαcos(π/3+α)-sin²(π/6-α)

题目

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°

答案

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