α-β∈(π/2,π) cos(α-β)0
cos(α+β)=√[1-sin²(α+β)]=√[1-(-12/13)²]=5/13
cos(2α)=cos[(α+β)+(α-β)]
=cos(α+β)cos(α-β)-sin(α+β)sin(α-β)
=(5/13)(-3/5)-(-12/13)(4/5)
=33/65
cos(2β)=cos[(α+β)-(α-β)]
=cos(α+β)cos(α-β)+sin(α+β)sin(α-β)
=(5/13)(-3/5)+(-12/13)(4/5)
=-63/65?????sin2????cos2??sin(2??)=sin[(??+??)+(??-??)]=sin(??+??)cos(??-??)+cos(??+??)sin(??-??)=(-12/13)(-3/5)+(5/13)(4/5)=56/65