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设a属于(π/4,3π/4),b属于(0,π/4）,cos(a-π/4)=3/5,sin(3π/4+b）=5/13,求sin（a+b）的值.
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设a属于(π/4,3π/4),b属于(0,π/4）,cos(a-π/4)=3/5,sin(3π/4+b）=5/13,求sin（a+b）的值.
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答案和解析
∵cos(a-π/4)=3/5, ∴cos(π/4-a)=3/5,
∵a∈(π/4,3π/4), ∴-π/2＜π/4-a＜0,
∴sin(π/4-a)=-4/5.
∵sin(3π/4+b）=5/13,∴sin(π/4-b）=5/13,
∵b∈(0,π/4）,∴0＜π/4-b＜π/4,
∴cos(π/4-b）=12/13.
∴sin（a+b）=cos[(π/2)-a-b]
=cos[(π/4-a)+(π/4-b)]
=cos(π/4-a)·cos(π/4-b)-sin(π/4-a)·sin(π/4-b)
=(3/5)×(12/13)-(-4/5)×(5/13)
=56/65.
∵a∈(π/4,3π/4), ∴-π/2＜π/4-a＜0,
∴sin(π/4-a)=-4/5.
∵sin(3π/4+b）=5/13,∴sin(π/4-b）=5/13,
∵b∈(0,π/4）,∴0＜π/4-b＜π/4,
∴cos(π/4-b）=12/13.
∴sin（a+b）=cos[(π/2)-a-b]
=cos[(π/4-a)+(π/4-b)]
=cos(π/4-a)·cos(π/4-b)-sin(π/4-a)·sin(π/4-b)
=(3/5)×(12/13)-(-4/5)×(5/13)
=56/65.
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