数学三角函数，两角的和与差的余弦。谢谢！

因为0<α<π/2<β<π，
所以π/3<π/3+α<π/3+π/2<π， -π/2<2π/3-π<2π/3-β<2π/3-π/2<π/2，
故由cos（π/3+α）=-3/5，可得sin（π/3+α）=4/5，
由sin（2π/3-β）=5/13，可得cos（2π/3-β）=12/13，
因此 cos（β-α）= cos（α-β）= -cos（π+α-β）
=-cos（(π/3+α)+(2π/3-β)）
=-(cos(π/3+α)cos(2π/3-β)-sin(π/3+α)sin(2π/3-β))
=-( (-3/5)*(12/13) - (4/5)*(5/13) )
=56/65

这个可以用简便方法的：∴ ∵α β
∵sin(2π/3-β)=sin(π-π/3-β)=sin(π/3+β)
∴cos(β-α)=cos[(π/3+β)-(π/3+α)]=cos(π/3+β)cos(π/3+α)+sin(π/3+β)sin(π/3+α)
∵0<α<π/2<β<π，cos（π/3+α）=-3/5，sin（2π/3-β）=5/13
∴cos(π/3+β)=-12/13
cos(π/3+α)=-3/5
sin(π/3+β)=5/13
sin(π/3+α)=4/5
∴原式=cos(β-α)=cos[(π/3+β)-(π/3+α)]=cos(π/3+β)cos(π/3+α)+sin(π/3+β)sin(π/3+α)=56/65
打的真累- -#