Question

已知cos(a-2/b)=12/13,sin(b-a/2)=4/5,且a属于(π/2,π),b(0,π/2),求sin(

Answer 1

因为a属于（π/2π）,b属于（0,π/2）

所以sin(a-b/2)=5/13,cos(b-a/2)=3/5

sin(a/2+b/2)=sin[(a-b/2)+(b-a/2)]

=sin(b-a/2)*cos(a-b/2)+sin(a-b/2)*cos(b-a/2)=63/65