已知α,β∈(3π/4,π),cos(α+β)=4/5,sin(β-π/4)=12/13,求cos(α+π/4)
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已知α,β∈(3π/4,π),cos(α+β)=4/5,sin(β-π/4)=12/13,求cos(α+π/4)
已知α,β∈(3π/4,π),cos(α+β)=4/5,sin(β-π/4)=12/13,求cos(α+π/4)
已知α,β∈(3π/4,π),cos(α+β)=4/5,sin(β-π/4)=12/13,求cos(α+π/4)
cos(α+β)=cos((α+π/4)+(β-π/4))=cos(α+π/4)cos(β-π/4)-SIN(α+π/4)SIN(β-π/4)=4/5
sin(α+β)=sin((α+π/4)+(β-π/4))=sin(α+π/4)cos(β-π/4)+cos(α+π/4)SIN(β-π/4)=-3/5
(IV象限)
COS(β-π/4)=-5/13(II象限)
cos(α+π/4)(-5/13)-SIN(α+π/4)(12/13)=4/5
sin(α+π/4)(-5/13)+cos(α+π/4)(12/13)=-3/5
cos(α+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)
α,β∈(3π/4,π),则α+β∈(3π/2,2π),β-π/4∈(π/2,3π/4),
所以sin(α+β)<0 cos(β-π/4)<0
而cos(α+β)=4/5,sin(β-π/4)=12/13
则sin(α+β)=...
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cos(α+π/4)=cos[(α+β)+(β-π/4)]=cos(α+β)×cos(β-π/4)-sin(α+β)×sin(β-π/4)
α,β∈(3π/4,π),则α+β∈(3π/2,2π),β-π/4∈(π/2,3π/4),
所以sin(α+β)<0 cos(β-π/4)<0
而cos(α+β)=4/5,sin(β-π/4)=12/13
则sin(α+β)=3/5 cos(β-π/4)=-5/13
cos(α+π/4)=cos[(α+β)+(β-π/4)]=cos(α+β)×cos(β-π/4)-sin(α+β)×sin(β-π/4)
=4/5×(-5/13)-3/5×12/13=-56/65
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