已知cos(π/4-α)=3/5,sin(5π/4+β)=-12/13,且β∈（0,π/4）,α∈（π/4,3π/4),求sin(α+β)

题目

已知cos(π/4-α)=3/5,sin(5π/4+β)=-12/13,且β∈（0,π/4）,α∈（π/4,3π/4),求sin(α+β)

答案

cos(π/4-α)=3/5
0<α<π/4
0<π/4-α<π/4
sin(π/4-α)>0
sin²(π/4-α)+cos²(π/4-α)=1
sin(π/4-α)=4/5
sin(π+π/4+β)=-12/13
sin(π/4+β)=12/13
π/4<β<3π/4
π/2<π/4+β<π
第二象限
cos(π/4+β)<0
sin²(π/4+β)+cos²(π/4+β)=1
所以cos(π/4+β)=-5/13
sin(α+β)
=sin[(π/4+β)-(π/4-α)]
=sin(π/4+β)cos(π/4-α)-cos(π/4+β)sin(π/4-α)]
=44/65