Question

已知cos(α+β)=5/13,sin(β-π/4)=-4/5,α,β∈(0,π/2),求cos2α

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Answer 1

∵α+β∈(0, π) ∴sin(α+β)=12/13
∵β-π/4∈(-π/4, π/4) ∴sin(α+β)=3/5

sin(α+π/4)=sin[(α+β)-(β-π/4)]=sin(α+β)cos(β-π/4)-cos(α+β)sin(β-π/4)=56/65
cos(α+π/4)=cos[(α+β)-(β-π/4)]=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)=-33/65

∴sin(2α+π/2)=2sin(α+π/4)cos(α+π/4)=-3696/4225
∴cos(2α)=-3696/4225