解:由(4cosa- 3sina )(2cosa- 3sina)=0 得 4cosa- 3sina=0或2cosa- 3sina=0 即tana=4/3或2/3 又角a∈(π/4,π/2 ) 则tana∈(1,+∞) 故tana=4/3 则cosa=3/5,sina=4/5 tan( a+π/4)=(tana+tanπ/4)/(1-tana.tanπ/4 )=-7 cos(π/3-2a)=cosπ/3.cos2a +sinπ/3.sin2a =(cosa.cosa-sina.sina)/2 + √3 sina.cosa =(24√3 - 7)/50