已知cos[a-(b⼀2)]=-1⼀9,sin[(a⼀2)-b]=2⼀3.且π⼀2<a<π,0<b<π⼀2,求sin(a+b)⼀2

2025-01-05 17:09:24
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回答(1):

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∴π/2<α<π,0<β<π/2, sinα=4√5/9,cosβ=√5/3.
∴sin(a+b)/2=4√5/9*√5/3-(-1/9)*2/3=22/27