Question

设cos(a-b/2)=-1/9,sin(a/2-b)=2/3,其中a属于(π/2,π),b属于(0,π/2),求cos

Answer 1

由sinA-cosA=1得sin(a-b/2)=九分之四倍根号5,cos(a/2-b)=3分之根号5

由cos(a-b/2)*cos(a/2-b)-sin(a-b/2)*sin(a/2-b)=cos（a/2+b/2）得cos（a/2+b/2）=负3分之根号5

由cos2A=2cosA^2-1得cos(a+b) =1/9