Question

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

Answer 1

已知π/4<β<α<3π/4
所以0<α-β<π/2
π/2<α+β<3π/2
cos(α-β)=12/13,sin(α+β)=-3/5
所以sin(α-β)=√[1-(12/13)²]=5/13
cos(α+β)=-√[1-(-3/4)²]=-4/5

所以sin2α=sin[(α-β)+(α+β)]
=sin(α-β)cos(α+β)+cos(α-β)sin(α+β)
=(5/13)*(-4/5)+(12/13)*(-3/5)=-56/65

Answer 2

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