已知α属于(π/4,3π/4),β属于(0,π/4),且cos(π/4-α)=3/5,sin(5π/4+β)=-12/13求cos(α+β)
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已知α属于(π/4,3π/4),β属于(0,π/4),且cos(π/4-α)=3/5,sin(5π/4+β)=-12/13求cos(α+β)
已知α属于(π/4,3π/4),β属于(0,π/4),且cos(π/4-α)=3/5,sin(5π/4+β)=-12/13
求cos(α+β)
已知α属于(π/4,3π/4),β属于(0,π/4),且cos(π/4-α)=3/5,sin(5π/4+β)=-12/13求cos(α+β)
α∈(π/4,3π/4),
α-π/4∈(0,π/2),
cos(π/4-α)=3/5
cos(α-π/4)=3/5
sin(α-π/4)=4/5
β∈(0,π/4)
π/4+β∈(π/4,π/2)
sin(π+π/4+β)=-12/13
-sin(π/4+β)=-12/13
sin(π/4+β)=12/13
cos(π/4+β)=5/13
cos(α+β)
=cos[(α-π/4)+(π/4+β)]
=cos(α-π/4)cos(π/4+β)-sin(α-π/4)sin(π/4+β)
=3/5*5/13-4/5*12/13
=15/65-48/65
=-33/65
呼~~~好难算。。。。答案是负65分之16
已知α属于(π/4,3π/4),β属于(0,π/4),
则-α属于(-3π/4,-π/4),π/4+β属于(0,π/2),π/4-α属于(-π/2,0),
cos(π/4-α)=3/5,0>sin(π/4-α)=-√[1-cos²(π/4-α)]=-4/5,sin(5π/4+β)=-sin(π/4+β)=-12/13,
sin(π/4+β)=12/13,0
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已知α属于(π/4,3π/4),β属于(0,π/4),
则-α属于(-3π/4,-π/4),π/4+β属于(0,π/2),π/4-α属于(-π/2,0),
cos(π/4-α)=3/5,0>sin(π/4-α)=-√[1-cos²(π/4-α)]=-4/5,sin(5π/4+β)=-sin(π/4+β)=-12/13,
sin(π/4+β)=12/13,0
=(5/13)(3/5)+(12/13)(-4/5)=-33/65
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