Question

已知2分之派<b<a<4分之3派，cos(a-b)=13分之12，sin(a+b)=-5分之3，则sin2a=

Answer 1

2a=(a+b)+(a-b)
π/2<b<a<3π/4
0<a-b<π/4 π<a+b<3π/2
cos(a-b)=12/13 ,sin(a-b)=5/13
sin(a+b)=-3/5 , cos(a+b)=-4/5
sin2a=sin[(a+b)+(a-b)]
=sin(a+b)cos(a-b)+cos(a+b)sin(a-b)
=-3/5 X 12/13+(-4/5) X 5/13
=-56/65

Answer 2

因为：2a=(a+b)+(a-b)
π/2<b<a<3π/4
所以：0<a-b<π/4 π<a+b<3π/2
cos(a-b)=12/13 ,sin(a-b)=5/13
sin(a+b)=-3/5 , cos(a+b)=-4/5
sin2a=sin[(a+b)+(a-b)]
=sin(a+b)cos(a-b)+cos(a+b)sin(a-b)
=-3/5 X 12/13+(-4/5) X 5/13
=-56/65