已知cos(α+β)=
|
∵α,β∈(0,
),∴α+β∈(0,π)π 2
∴sin(α+β)=
=1-co s2(α+β)
=1-(
)25 13 12 13
∴sinβ=
=1-cos2β
=1-(
)24 5 3 5
cosα=cos[(α+β)-β]=cos(α+β)cosβ+sin(α+β)sinβ
=
×5 13
+4 5
×12 13 3 5
=56 65
sin(α+2β)=sin[(α+β)+β]=sin(α+β)cosβ+cos(α+β)sinβ
=
×12 13
+ 4 5
× 5 13 3 5
=63 65