设cos(a-1/2b)=-1/9,sin(1/2a-b)=2/3,且π/2
设cos(a-1/2b)=-1/9,sin(1/2a-b)=2/3,且π/2
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
设cos(A-B/2)=-1/9,sin(A/2-B)=2/3,且A是第二象限角B是第一象限角求cos(A+B)/2的值
已知cos(a-B/2)=-1/9,sin(a/2-B)=2/3,且π/2
cos(a-b/2)=-1/9 sin(a/2-b)=2/3 且π/2
若sin^4a/sin^2b+cos^4a/cos^2b=1,证明sin^4b/sin^2a+cos^4b/cos^2a
已知cos(a-b/2)=-1/9,sin(a/2-b)=2/3,且a∈(∏/2,∏),b∈(0,∏/2),则cos[(
已知cos[a-(b/2)]=-1/9,sin[(a/2)-b]=2/3.且π/2
已知cos(a-B/2)=-1/9,sin(a/2-B)=2/3,且0
已知cos(a-b\2)=-1\9,sin(a\2-b)=2\3,且0
cos(A+B)cosB+sin(A+B)sinB=1/3,且A∈(3π/2,2π),求cos(2A+(π/4))
cos(a+B)×cos(a-B)=1/3,求cos^2(a)-sin^2(B)的值