已知sin(a+3TT4)=5/13,cos(TT/4-β)=3/5,且-TT/4

1个回答

  • ∵﹣π/4<α<π/4 ∴π/2<α+3π/4<π

    ∵sin(α+3π/4)=5/13 ∴cos(α+3π/4)=﹣12/13

    ∵π/4<β<3π/4 ∴﹣3π/4<﹣β<﹣π/4 ∴﹣π/2<π/4-β<0

    ∵cos(π/4-β)=3/5 ∴sin(π/4-β)=﹣4/5

    ∵cos[(α+3π/4)+(π/4-β)]=cos[π+(α-β)]=﹣cos(α-β)

    ∵cos[(α+3π/4)+(π/4-β)]=cos(α+3π/4)cos(π/4-β)-sin(α+3π/4)sin(π/4-β)

    =(﹣12/13)×3/5-5/13×(﹣4/5)=﹣16/65

    ∴cos(α-β) =16/65

    ∴cos2(α-β)=2 cos²(α-β) -1=2×(16/65)²-1=