已知π/3<α<3π/4,sin(α-π/4)=3/5,求(sinα-cos2α+1)/tanα的值
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解答:
∵ π/3<α<3π/4
∴ π/12<α-π/4<π/2
∵ sin(α-π/4)=3/5
∴ cos(α-π/4)=√[1-sin²(α-π/4)]=√(1-9/25)=4/5
∴ sinα=sin[(α-π/4)+π/4]
=sin(α-π/4)cos(π/4)+cos(α-π/4)sin(π/4)
=(3/5)*(√2/2)+(4/5)*(√2/2)
=7√2/10
cosα=cos[(α-π/4)+π/4]
=cos(α-π/4)cos(π/4)-sin(α-π/4)sin(π/4)
=(4/5)*(√2/2)-(3/5)*(√2/2)
=√2/10
∴(sinα-cos2α+1)/tanα
=(sinα+2sin²α)/(sinα/cosα)
=sinα(1+2sinα)/(sinα/cosα)
=cosα(1+2sinα)
=(√2/10)*(1+7√2/5)
=(5√2+14)/50
∵ π/3<α<3π/4
∴ π/12<α-π/4<π/2
∵ sin(α-π/4)=3/5
∴ cos(α-π/4)=√[1-sin²(α-π/4)]=√(1-9/25)=4/5
∴ sinα=sin[(α-π/4)+π/4]
=sin(α-π/4)cos(π/4)+cos(α-π/4)sin(π/4)
=(3/5)*(√2/2)+(4/5)*(√2/2)
=7√2/10
cosα=cos[(α-π/4)+π/4]
=cos(α-π/4)cos(π/4)-sin(α-π/4)sin(π/4)
=(4/5)*(√2/2)-(3/5)*(√2/2)
=√2/10
∴(sinα-cos2α+1)/tanα
=(sinα+2sin²α)/(sinα/cosα)
=sinα(1+2sinα)/(sinα/cosα)
=cosα(1+2sinα)
=(√2/10)*(1+7√2/5)
=(5√2+14)/50
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