Question

已知0<α<π/2＜β＜π, sinα=3/5 ,cos(β-α)=√2/10,求2α-β

已知0<α<π/2＜β＜π, sinα=3/5 ,cos(β-α)=√2/10,求2α-β

Answer 1

解：∵0<α<π/2＜β＜π
∴0<β-α<π，-π<2α-β<π/2
∴cosα>0，sin(β-α)>0
∵sinα=3/5 ,cos(β-α)=√2/10
∴cosα=√(1-sin²α)=4/5,
sin(β-α)=√(1-cos²(β-α))=7√2/10
∴sin(2α-β)=sin(α-(β-α))
=sinαcos(β-α)-cosαsin(β-α)
=(3/5)(√2/10)-(4/5)(7√2/10)
=-√2/2
即2α-β=-π/4，或2α-β=-3π/4 (∵-π<2α-β<π/2)

Answer 2

cosa=4/5;sin(β-α)=7√2 /10;
sin(2α-β)=sin(α-(β-α))
=(3/5 )*(√2/10)-(7√2/10)*(4/5)
=-√2/2;
2α-β=-25°