tan(α+β)=2/5 ,tan(β-π4 )=1/4 ,则sin(α+π4 )•sin(π4 -α)=
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tan(α+β)=2/5 ,tan(β-π4 )=1/4 ,则sin(α+π4 )•sin(π4 -α)=
tan(α+β)=2/5 ,tan(β-π4 )=1/4 ,则sin(α+π4 )•sin(π4 -α)=
tan(α+β)=2/5 ,tan(β-π4 )=1/4 ,则sin(α+π4 )•sin(π4 -α)=
解析:
已知tan(α+β)=2/5 ,tan(β-π4 )=1/4,那么:
tan(α+π/4)=tan[(α+β)-(β-π4 )]
=[tan(α+β)- tan(β-π4 )]/[1+tan(α+β)tan(β-π4 )]
=(2/5 - 1/4)/(1+2/5 * 1/4)
=(3/20)/(3/20)
=1
又tan(α+π/4)=sin(α+π/4)/cos(α+π/4),那么:
sin(α+π/4)=cos(α+π/4)
而sin²(α+π/4)+cos²(α+π/4)=1
易得:sin²(α+π/4)=1/2
所以:sin(α+π/4 )•sin(π/4 -α)
=sin(α+π/4 )•sin(π/2- π/4 -α)
=sin(α+π/4)cos(α+π/4)
=sin²(α+π/4)
=1/2
原式=0.5