已知cos[a-(b/2)]=-1/9,sin[(a/2)-b]=2/3.且π/2
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![已知cos[a-(b/2)]=-1/9,sin[(a/2)-b]=2/3.且π/2](/uploads/image/z/1730032-16-2.jpg?t=%E5%B7%B2%E7%9F%A5cos%5Ba-%28b%2F2%29%5D%3D-1%2F9%2Csin%5B%28a%2F2%29-b%5D%3D2%2F3.%E4%B8%94%CF%80%2F2)
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π/2
已知cos[a-(b/2)]=-1/9,sin[(a/2)-b]=2/3.且π/2
设a-(b/2)=α ,(a/2)-b=β
则(a+b)/2=α-β
∴sin(a+b)/2=sin(α-β)=sinαcosβ-cosαsinβ
∵π/2