It is known that cos (2 α - β) = - 11 / 14, sin (α - 2 β) = 4 √ 3 / 7, 0 ＜ β ＜ π / 4 ＜ α ＜ π / 2

It is known that cos (2 α - β) = - 11 / 14, sin (α - 2 β) = 4 √ 3 / 7, 0 ＜ β ＜ π / 4 ＜ α ＜ π / 2

cos(α+β)=cos((2α-β)-(α-2β))=cos(2α-β)cos(α-2β)-sin(α-2β)sin(2α-β)=-11/14×1/7+4√3/7×5√3/14=1/2
so α+β=60°

It is known that cos (2 α - β) = - 11 / 14, sin (α - 2 β) = (4 √ 3) / 7, π / 4

cos(α+β)= cos[(2α-β)-(α-2β)]=cos(2α-β)cos(α-2β)+sin(2α-β)sin(α-2β)
∵π/4

Given π / 2 〈α〈π, π / 2 〈β〈π, sin α = 3 / 5, cos β = - 12 / 13, find the value of COS (α - β)

α+β∈(π/2,π)
<=>cos(α+β)<0
Cos ^ (α + β) + sin ^ (α + β) = 1, sin (α + β) = 3 / 5
Cos (α + β) = - 4 / 5 is obtained
Similarly:
α-β∈(π,3π/2)
<=>sin(α-β)<0
From sin ^ (α - β) + cos ^ (α - β) = 1, cos (α - β) = - 5 / 13
The results show that sin (α - β) = - 12 / 13
therefore:
sin2α=sin[(α+β)+(α-β)]=sin(α+β)*cos(α-β)+cos(α+β)*sin(α-β)
=(3/5)*(-5/13)+(-4/5)*(-12/13)
=33/65