已知:0<α<π/2<β<π,cos(β-π/4)=1/3,sin(α+β)=4/5 (1)求sin2β (2)求cos
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已知:0<α<π/2<β<π,cos(β-π/4)=1/3,sin(α+β)=4/5 (1)求sin2β (2)求cos(α+π/4)的值
![已知:0<α<π/2<β<π,cos(β-π/4)=1/3,sin(α+β)=4/5 (1)求sin2β (2)求cos](/uploads/image/z/15986211-51-1.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%9A0%EF%BC%9C%CE%B1%EF%BC%9C%CF%80%2F2%EF%BC%9C%CE%B2%EF%BC%9C%CF%80%2Ccos%28%CE%B2-%CF%80%2F4%29%3D1%2F3%2Csin%28%CE%B1%2B%CE%B2%29%3D4%2F5+%281%29%E6%B1%82sin2%CE%B2+%282%29%E6%B1%82cos)
cos(β-π/4)=1/3
cos2(β-π/4)=2[cos(β-π/4)]^2-1=-7/9
cos(2β-π/2)=-7/9
cos(π/2-2β)=-7/9
sin2β=-7/9
cos(β-π/4)=1/3,sin(α+β)=4/5,0<α<π/2<β<π
sin(β-π/4)=2√2/3
cos(α+β)=-3/5
cos(α+π/4)=cos[(α+β)+(π/4-β)]
=cos(α+β)cos(π/4-β)-sin(α+β)sin(π/4-β)
=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)
=(8√2-3)/15
cos2(β-π/4)=2[cos(β-π/4)]^2-1=-7/9
cos(2β-π/2)=-7/9
cos(π/2-2β)=-7/9
sin2β=-7/9
cos(β-π/4)=1/3,sin(α+β)=4/5,0<α<π/2<β<π
sin(β-π/4)=2√2/3
cos(α+β)=-3/5
cos(α+π/4)=cos[(α+β)+(π/4-β)]
=cos(α+β)cos(π/4-β)-sin(α+β)sin(π/4-β)
=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)
=(8√2-3)/15
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