若sin(π/6-a)cos(π/3+b)+sin(π/3+a)cos(π/6-b)=12/13,sin(a+b)=-3

已知sin[a-b]cos a-cos[b-a]sin a=3/5,b是第三象限角,求sin[b+5π/4]的值 已知sin(a+b)=-3/5,cos(a-b)=12/13,且π/2 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) 已知cos(a+b)=0,求值:sin(π+2a+b)-cos(-π/2+2a+3b) 已知a,b属于(3π/4,π),sin(a+b)=-3/5,sin(b-π/4)=12/13,则cos(a+4/π)= 已知A B属于3π/4到π,SIN(A+B)=-3/5,SIN(B-π/4)=12/13,求COS(A+π/4 已知向量a=(sin(α+π/6),1),b=(4,4cosα-根号3),若a⊥b,则sin(α+4π/3)等于 已知向量a=(cosα,sinα),向量b=(cosβ,sinβ) 若α-β=π/3,求a+2b向量的绝对值 化简：sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5 sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5则si 已知角ab sin(a+b)=-3/5 sin(b-π/4)=12/13 求cos(a+π/4)) 答案是-56/65 已知角ab sin(a+b)=-3/5 sin(b-π/4)=12/13 求cos(a+π/4))