作业帮已知-π/4＜α＜π/4,π/4＜β＜3π/4,sin（3π/4+α）=5/13,cos(π/4-β）=3/5,求sin2(α-β）具体解答过程
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已知-π/4＜α＜π/4,π/4＜β＜3π/4,sin（3π/4+α）=5/13,cos(π/4-β）=3/5,求sin2(α-β）具体解答过程
已知-π/4＜α＜π/4,π/4＜β＜3π/4,sin（3π/4+α）=5/13,cos(π/4-β）=3/5,求sin2(α-β）
具体解答过程
已知-π/4＜α＜π/4,π/4＜β＜3π/4,sin（3π/4+α）=5/13,cos(π/4-β）=3/5,求sin2(α-β）具体解答过程
∵-π/4＜α＜π/4,π/4＜β＜3π/4
==>π/2＜3π/4+α＜π,-π/2＜π/4-β＜0
==>α+3π/4是第二象限的角,π/4-β是第四象限的角
又sin(3π/4+α)=5/13,cos(π/4-β)=3/5
∴cos(3π/4+α)=-√[1-sin²(3π/4+α)]=-12/13,sin(π/4-β)=-√[1-cos²(π/4-β)]=-4/5
故sin[2(α-β)]=sin{2[(3π/4+α)+(π/4-β)-π]}
=-sin{2[π-(3π/4+α)-(π/4-β)]}
=-sin{2π-2[(3π/4+α)+(π/4-β)]}
=sin{2[(3π/4+α)+(π/4-β)]}
=2sin[(3π/4+α)+(π/4-β)]*cos[(3π/4+α)+(π/4-β)]
=2[sin(3π/4+α)*cos(π/4-β)+cos(3π/4+α)*sin(π/4-β)]*[cos(3π/4+α)*cos(π/4-β)
-sin(3π/4+α)*sin(π/4-β)]
=2[(5/13)*(3/5)+(-12/13)*(-4/5)]*[(-12/13)*(3/5)-(5/13)*(-4/5)]
=2(63/65)(-56/65)
=-7056/4225