tana=1/2,求tan2a和sin（2a+派/3）

由已知可得：
sin(2a+ π/3)=sin[π/2 + (2a-π/6)]=cos(2a - π/6)=1/3
则有：2cos平方(a- π/12) - 1=1/3即得：cos平方(a- π/12) =2/3
又a属于(0,π/2),那么：a- π/12属于(- π/12,5π/12)
可知：cos(a- π/12)>0
所以解cos平方(a- π/12) =2/3得：cos(a- π/12)=(根号6)/3

f(x)=根号3sinx+cosx
=2(√3/2sinx+1/2cosx）
=2(sinxcosπ/6+cosxsinπ/6)
=2sin(x+π/6)
所以,函数的最小正周期=2π；最大值=2
f(a)=2sin(a+π/6)=2/3
sin(a+π/6)=1/3
cos(2a+π/3)=cos[2(a+π/6)]
=1-2sin²(a+π/6)
=1-2×1/9
=1-2/9
=7/9