若sinA+cosA/sinA-cosA=2,则sin(A-5π)*sin(7π/2-A)等于?

sinA+cosA/sinA-cosA=2
tanA=3
sin(A-5π)*sin(7π/2-A)
=-sinA*sin[4π-(π/2+A)]
=sinA*sin(π/2+A)
=sinAcosA
=1/2sin2A
=1/2*2tanA/(1+tan^2A)
=tanA/(1+tan^2A)
=3/(1+9)
=3/10

(sinA+cosA)/(sinA-cosA)=2
sinA+cosA=2sinA-2cosA
3cosA=sinA
sin²A+cos²A=1，9cos²A+cos²A=1，cos²A=1/10
sin(A-5π)*sin(7π/2-A)
=(-sin(5π-A)sin(2π+3π/2-A)
=-sin(4π+π-A)sin(3π/2-A)
=-sin(π-A)(-cosA)
=sinAcosA
=3cosAcosA
=3cos²A
=3/10

（sina+cosa)/(sina-cosa)=2所以sina=3cosa因为sina*sina+cosa*cosa=1,所以sina=3*根号10/10,cosa=根号10/10,或者sina=-3*根号10/10,cosa=-根号10/10；sin(A-5π)*sin(7π/2-A)=(-sinA)*(-sin(A+π/2)=sina*cosa=3/1...