设cos(a-b/2)=-1/9 sin(a/2-b)=2/3 a∈(π/2,π) b∈(0,π/2) 求sin(a+b
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
高一数学已知sin(π/2-b)*cos(a+b)-sin(π+b)*sin(a+b)=3/5其中a∈(3π/2,2π)
设cos(a-1/2b)=-1/9,sin(1/2a-b)=2/3,且π/2
已知sin(A+B)=4/5,sinA=3/5,A,B∈[0,π/2],(1)求cos(A+B)的值 (2)求cosB的
已知cos(a+b)=0,求值:sin(π+2a+b)-cos(-π/2+2a+3b)
若sin^4a/sin^2b+cos^4a/cos^2b=1,证明sin^4b/sin^2a+cos^4b/cos^2a
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
设a>0,0<b<π/2,且a+b=5π/6,求函数y=2-sin^a - cos^b的值域
已知0<A<π/2<B<π,cos(B-π/4)=1/3,sin(A+B)=4/5求cos(a+
Cos(a+b)*cos(a-b)=1/5 求cos ^2-sin^2
设向量a=(cos(x/2),sin(x/2)),向量b=(sin(3x/2),cos(3x/2)),x∈[0,π/2]