已知sin(π/6-α)=1/5,且α∈(0,π/2),则cosα=

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已知sin(π/6-α)=1/5,且α∈(0,π/2),则cosα=已知sin(π/6-α)=1/5,且α∈(0,π/2),则cosα=已知sin(π/6-α)=1/5,且α∈(0,π/2),则cos

已知sin(π/6-α)=1/5,且α∈(0,π/2),则cosα=
已知sin(π/6-α)=1/5,且α∈(0,π/2),则cosα=

已知sin(π/6-α)=1/5,且α∈(0,π/2),则cosα=
解α∈(0,π/2)
知-α∈(-π/2,0)
π/6-α∈(-π/3,π/6)
由sin(π/6-α)=1/5
知π/6-α是第一象限角
即cos(π/6-α)=2√6/5
cosα
=cos-α
=cos[(π/6-α)-π/6]
=cos(π/6-α)cos(π/6)+sin(π/6-α)sin(π/6)
=2√6/5×√3/2+1/5×1/2
=(6√2+1)/10


α∈(0,π/2)
π/6-α∈(-π/3,π/6)
∴ cos(π/6-α)>0
∵ cos²(π/6-α)=1-sin²(π/6-α)=1-1/25=24/25
∴ cos(π/6-α)=2√6/5
∴ cosα
=cos[π/6-(π/6-α)]
=cos(π/6)cos(π/6-α)] +sin(π/6)sin(π/6-α)
=(√3/2)*(2√6)/5+(1/2)*(1/5)
=(6√2+1)/10