1)若tana=1/3,则cosa+1/2sin2a 解: tana=1/9 sina/cosa=1/9 又sina+cosa=1 所以cosa=9/10 cosa+1/2*2sinacosa =cosa+sinacosa =cosa(1+sina/cosa) =cosa(1+tana) =9/10*(1+1/3) =1.2 2)函数y=cosx+cos(x+π/3)的最大值为__ 解:先化简 y=cosx+cos(x+π/3) =cosx+cosxcosπ/3-sinxsinπ/3 =cosx+(1/2)cosx-(√3/2)sinx =(3/2)cosx-(√3/2)sinx =√3[(√3/2)cosx-(1/2)sinx] =√3(sinπ/3*cosx-cosπ/3*sinx) =√3sin(π/3-x) 很明显,最大值为√3 3)sinα(1+cotα)+cosα(1+tanα)=sinα+cosα 证明上面三角恒等式相等 证明:sinα(1+cotα)+cosα(1+tanα) =sinα(1+cosα/sinα)+cosα(1+sinα/cosα) =sinα(sinα+cosα)+cosα(sinα+cosα) =(sinα+cosα)(sinα+cosα) =sinα+cosα (不算难的,希望对你有所帮助哈!)