已知cosa(π/6-α)=根号3/3 ,求sin(π/2+α)
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已知cosa(π/6-α)=根号3/3 ,求sin(π/2+α)
已知cosa(π/6-α)=根号3/3 ,求sin(π/2+α)
已知cosa(π/6-α)=根号3/3 ,求sin(π/2+α)
已知 cosa(π/6-α)=√3 / 3
求sin(π/2+α)
方法1：cos(α - β) = cosαcosβ + sinα sinβ
cosa(π/6-α) = cosπ/6*cosα + sinπ/6* sinα =√3 / 2*cosα + 1/2* sinα = √3 / 3
所以：√3 / 2*cosα + 1/2* √(1 - cos²α) = √3 / 3
1 - cos²α = 4 (√3 / 3 - √3 / 2*cosα) ²
12cos²α - 12cosα +1 = 0
解得：cosα = [3 ±√6] / 6
答案：sin(π/2+α) = cosα ＝ [3 ±√6] / 6
cosa(π/6-α)=√3/3
sina(π/6-α)=±√6/3
当sina(π/6-α)=√6/3时
sin(π/2+α)
=sin[2π/3-(π/6-α)]
=sin(2π/3)cos(π/6-α)-cos(2π/3)sin(π/6-α)
=√3/2*√3/3-(-1/2)*√6/3
=1/2+√6/6
cosa(π/6-α)=√3/3
sina(π/6-α)=±√6/3
当sina(π/6-α)=√6/3时
sin(π/2+α)
=sin[2π/3-(π/6-α)]
=sin(2π/3)cos(π/6-α)-cos(2π/3)sin(π/6-α)
=√3/2*√3/3-(-1/2)*√6/3
=1/2+√6/6
当sina(π/6...
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cosa(π/6-α)=√3/3
sina(π/6-α)=±√6/3
当sina(π/6-α)=√6/3时
sin(π/2+α)
=sin[2π/3-(π/6-α)]
=sin(2π/3)cos(π/6-α)-cos(2π/3)sin(π/6-α)
=√3/2*√3/3-(-1/2)*√6/3
=1/2+√6/6
当sina(π/6-α)=-√6/3时
sin(π/2+α)
=sin[2π/3-(π/6-α)]
=sin(2π/3)cos(π/6-α)-cos(2π/3)sin(π/6-α)
=√3/2*√3/3-(-1/2)*（-√6/3)
=1/2-√6/6
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