已知π/2<B<A<3π/4,cos(A-B)=12/13,sin(A+B)=-3/5,求sin2A

问题描述:

已知π/2<B<A<3π/4,cos(A-B)=12/13,sin(A+B)=-3/5,求sin2A

π/2π/2所以π所以cos(A+B)sin(A+B)=-3/5
所以cos(A+B)=-4/5
-3π/4π/2所以-π/4A>B,
所以0sin(A-B)>0
cos(A-B)=12/13
所以sin(A-B)=5/13
sin2A=sin[(A+B)+(A-B)]
=sin(A+B)cos(A-B)+cos(A+B)sin(A-B)
=-56/65