已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,求cos(α+π/4)
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已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,求cos(α+π/4)已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/
已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,求cos(α+π/4)
已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,求cos(α+π/4)
已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,求cos(α+π/4)
cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
=cos(a+b)cos(b-π/4)-36/65;
3π/4
3π/4
所以：
cos(a+π/4)=20/65-36/65=-16/65.
cos(α+π/4)
=cos[α+β-（β-π/4）]
=cos(α+β)cos(β-π/4)-sin(α+β)sin(β-π/4)
sin(α+β)=-3/5 α+β ∈(3π/2,2π), sina^2+cosa^2=1
所以cos(α+β)=4/5
β-π/4∈(π/2,3π/4)
所以cos(β-π/4)=-5/13
代入就行了
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