(1)由题意得(sinα+cosα)2= ,即1+sin2α= ,∴sin2α= . 又2α∈(0, ),∴cos2α= = ,∴tan2α= = . ……4分 (2)∵β∈( , ),β- ∈(0, ),∴cos(β- )= , 于是sin2(β- )=2sin(β- )cos(β- )= . 又sin2(β- )=-cos2β,∴cos2β=- .又2β∈( ,π),∴sin2β= . 又cos2α= = ,∴cosα= ,sinα= (α∈(0, )). ∴cos(α+2β)=cosαcos2β-sinαsin2β= ×(- )- × =- . |