Question

已知cos(a-b/2)=12/13,sin(b-a/2)=4/5,a属于(90,180),b属于(0,90),求(a/2+b/2)

rt..过程,,谢谢

Answer 1

因为a属于（π/2，π），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