Question

已知0<α<π/4<β<3π/4,cos(π/4 -β)=3/5,sin(3π/4+α)=5/13,求sin(α+β)的值

详细过程，结果请求出 谢谢

Answer 1

解：∵0<α<π/4<β<3π/4
==>-π/2<π/4-β<0,3π/4<3π/4+α<π
∴sin(π/4-β)<0,cos(3π/4+α)<0
∵cos(π/4-β)=3/5,sin(3π/4+α)=5/13
∴sin(π/4-β)=-√[1-(cos(π/4-β))^2]=-4/5
cos(3π/4+α)=-√[1-(sin(3π/4+α))^2]=-12/13
故sin(α+β)=sin[(3π/4+α)-(π/4-β)-π/2]
=-sin[(3π/4+α)-(π/4-β)]
=cos(3π/4+α)sin(π/4-β)-sin(3π/4+α)cos(π/4-β) (应用差角公式)
=(-12/13)(-4/5)-(5/13)(3/5)
=33/65。