求证,sin²α+cosαcos(π/3+α)-sin²(π/6+α)的值与α无关
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/19 07:32:42
求证,sin²α+cosαcos(π/3+α)-sin²(π/6+α)的值与α无关
sin²α+cosαcos(π/3+α)-sin²(π/6-α)
=sin²α+cosαcos(π/3+α)-sin²[π/2-(π/3+α)]
=sin²α+cosαcos(π/3+α)-cos²(π/3+α)
=sin²α+cos(π/3+α)*[cosα-cos(π/3+α)]
=sin²α+cos(π/3+α)*[-2sin(-π/6)sin(π/6+α)]
=sin²α+cos(π/3+α)*sin(π/6+α)
=sin²α+cos(π/3+α)*cos(π/3-α)
=sin²α+[cos(π/3)cosα-sin(π/3)sinα]*[cos(π/3)cosα+sin(π/3)sinα]
=sin²α+cos²(π/3)cos²α-sin²(π/3)sin²α
=sin²α+1/4*cos²α- 3/4*sin²α
=1/4*sin²α+1/4*cos²α
=1/4
sin²α+cosαcos(π/3+α)-sin²(π/6-α)的值与α无关
=sin²α+cosαcos(π/3+α)-sin²[π/2-(π/3+α)]
=sin²α+cosαcos(π/3+α)-cos²(π/3+α)
=sin²α+cos(π/3+α)*[cosα-cos(π/3+α)]
=sin²α+cos(π/3+α)*[-2sin(-π/6)sin(π/6+α)]
=sin²α+cos(π/3+α)*sin(π/6+α)
=sin²α+cos(π/3+α)*cos(π/3-α)
=sin²α+[cos(π/3)cosα-sin(π/3)sinα]*[cos(π/3)cosα+sin(π/3)sinα]
=sin²α+cos²(π/3)cos²α-sin²(π/3)sin²α
=sin²α+1/4*cos²α- 3/4*sin²α
=1/4*sin²α+1/4*cos²α
=1/4
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