已知π/2﹤β﹤a﹤3π/4，cos(a-β)=12/13，sin(a+β)=-(3/5)，求sin2a的值．
因为π/2﹤β﹤3π/4,cos(a-β)=12/13,sin(a+β)=-(3/5)， 所以a-β∈(0,π/2)，a+β∈(π，3π/2)． 所以sin(a-β)=5/13，cos(a+β)=-(4/5)． sin2a=sin[(a+β)+(a-β)] =sin(a+β)cos(a-β)+cos(a+β)sin(a-β) =-(3/5)×1(2/13)-(4/5)×(5/13)=-(56/65).
