已知cos(π/4 -α)=3/5,sin(5π/4+β)=-12/13,α∈(π/4,3π/4),β∈(0,π/4),求sin(α+β)的值
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已知cos(π/4 -α)=3/5,sin(5π/4+β)=-12/13,α∈(π/4,3π/4),β∈(0,π/4),求sin(α+β)的值
已知cos(π/4 -α)=3/5,sin(5π/4+β)=-12/13,α∈(π/4,3π/4),β∈(0,π/4),求sin(α+β)的值
已知cos(π/4 -α)=3/5,sin(5π/4+β)=-12/13,α∈(π/4,3π/4),β∈(0,π/4),求sin(α+β)的值
cos(π/4 -α)=3/5
α∈(π/4,3π/4)
-α∈(-3π/4,-π/4)
π/4-α∈(-π/2,0)
sin(π/4 -α)=-4/5
sin(5π/4+β)=-12/13
sin(π+π/4+β)=-12/13
-sin(π/4+β)=-12/13
sin(π/4+β)=12/13
β∈(0,π/4)
π/4+β∈(π/4,π/2)
cos(π/4+β)=5/13
sin(β+α)
=sin(π/4+β-π/4+α)
=sin[(π/4+β)-(π/4 -α)]
=sin(π/4+β)cos(π/4 -α)-cos(π/4+β)sin(π/4 -α)
=12/13*3/5-5/13*(-4/5)
=36/65+20/65
=56/65
sin[﹙﹙5π﹚/4+β﹚-﹙π/4-α﹚]=sin﹙π+β+α﹚=﹣sin(α+β)
又cos(π/4 -α)=3/5 α∈(π/4,3π/4),∴﹣π/2<π/4-α<0 ∴sin(π/4 -α)=﹣4/5
sin(5π/4+β)=-12/13 β∈(0,π/4),∴cos(5π/4+β)=-5/13
∴sin(α+β)=﹣sin[﹙...
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sin[﹙﹙5π﹚/4+β﹚-﹙π/4-α﹚]=sin﹙π+β+α﹚=﹣sin(α+β)
又cos(π/4 -α)=3/5 α∈(π/4,3π/4),∴﹣π/2<π/4-α<0 ∴sin(π/4 -α)=﹣4/5
sin(5π/4+β)=-12/13 β∈(0,π/4),∴cos(5π/4+β)=-5/13
∴sin(α+β)=﹣sin[﹙﹙5π﹚/4+β﹚-﹙π/4-α﹚]=sin(π/4 -α)cos(π/4 -α)-sin(5π/4+β)cos(5π/4+β)=﹣12/25-60/169
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