已知π⼀2<β<α<3π⼀4,且cos( α-β)=12⼀13, cos(α+β)=-4⼀5,求sin2α

2024-11-18 14:17:30
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回答1:

因为π/2<β<α<3π/4
所以0<α-β<π/4,π<α+β<3π/2
cos( α-β)=12/13, cos(α+β)=-4/5,
所以sin( α-β)=5/13, sin(α+β)=-3/5
则sin2α=sin [(α+β)+(α-β)] =sin(α+β)cos( α-β)+cos(α+β)sin(α-β)
= - 56/65