已知π⼀4<a<β<π⼀2,且sin(a+β)=4⼀5,cos(a-β)=12⼀13,求sin2a

2025-05-22 02:58:39
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回答(1):

∵π/4∴cos(α+β)=-3/5,sin(α-β)=-5/13
sin2α=sin(a-β+α+β)
=sin(α-β)cos(α+β)+cos(α-β)sin(α+β)
=5/13 ×3/5+ 4/5 ×12/13=63/65