Question

已知0<β<π/4<α<3π/4,cos(π/4-α)=3/5

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

Answer 1

cos[(π/4-a)-(3π/4+β)]
=cos(-π/2-a-β)
=cos(π/2+a+β)
=sin(a+β)

sin(a+β)=(3/5)(12/13)+(5/13)(4/5)
=56/65

2.

cos(π/4-α)=3/5
π/4<α<3π/4 -3π/4<-α<-π/4
则-π/2<π/4-α<0
sin(π/4-α)<0
sin(π/4-α)=-√(1-(cos(π/4-α))^2)=-4/5

sin(3π/4+β)=5/13
0<β<π/4 3π/4<3π/4+β<π
cos(3π/4+β)<0
cos(3π/4+β)=-√(1-(sin(3π/4+β))^2)=-12/13

sin(α+β)=-cos(α+β+π/2)=-cos((3π/4+β)-(π/4-α))
=-[cos(3π/4+β)cos(π/4-α)+sin(3π/4+β)sin(π/4-α)]
=-[(-12/13)*3/5+5/13*(-4/5)]
=56/65

Answer 2

cos〔3π/4+β-(π/4-α)〕=cos(π/2+α+β)=-sin(α+β)
=cos(3π/4+β)cos(π/4-α)+sin(3π/4+β)sin(π/4-α)
0<β<π/4<α<3π/4
所以3π/4<β+3π/4<2π，-π/2<α<0
sin(3π/4+β)=5/13＞0所以3π/4<β+3π/4<π,
cos(3π/4+β)＜0 sin(π/4-α)＜0
=-12/13 =-4/5
cos(3π/4+β)cos(π/4-α)+sin(3π/4+β)sin(π/4-α)
=-36/65-20/65=-56/65=-sin(α+β)
所以sin(α+β)=56/65