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)
问题描述：
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
cos(π/4-α)=3/5
00sin(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/4π/2第二象限
cos(π/4+β)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