已知函数f(x)=[cos(x/2)]^2-[sin(x/2)]^2+sinx
问题描述：
已知函数f(x)=[cos(x/2)]^2-[sin(x/2)]^2+sinx
1)求f(x)的最小正周期
2)当x0属于(0,π/4)且f(x0)=4√2/5时,求f(x0+π/6)的值
答
f(x)=[cos(x/2)]^2-[sin(x/2)]^2+sinx
=cos[2*(x/2)]+sinx
=sinx+cosx
=√2sin(x+π/4)
所以T=2π/1=2π
f(x0)=√2sin(x+π/4)=4√2/5
sin(x+π/4)=4/5
x0属于(0,π/4)
所以x0+π/4属于(π/4,π/2)
所以cos(x0+π/4)>0
因为sin²(x0+π/4)+cos²(x0+π/4)=1
sin(x0+π/4)=4/5
所以cos(x0+π/4)=3/5
f(x0+π/6)=√2sin(x0+π/4+π/6)
=√2[sin(x0+π/4)cosπ/6+cos(x0+π/4)sinπ/6]
=7√6/10