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

2024-11-18 14:16:14
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回答1:

sin(a-b) = 5/13 ,cos(a+b)=4/5
sin2a=sin{(a+b)+(a-b)}
=sin(a+b)cos(a-b) + sin(a-b)cos(a+b)
= - (3/5 )(12/13) + (5/13)(4/5)
=-16/65