一道数学会考题：已知函数f(x)=sin(2x+π/3)+cos(2x-π/6)（1）将函数f(x)的解析式化为f(x)=Asin(ωx+φ) 的形式,并指出它的最小正周期（2）当x∈[- π/6,π/6]时,求f(x)的最小值和相应x的值
问题描述：
一道数学会考题：已知函数f(x)=sin(2x+π/3)+cos(2x-π/6)
（1）将函数f(x)的解析式化为f(x)=Asin(ωx+φ) 的形式,并指出它的最小正周期
（2）当x∈[- π/6,π/6]时,求f(x)的最小值和相应x的值
答
(1)f(x)=sin2xcosπ/3+sinπ/3cos2x+cos2xcosπ/6+sin2xsinπ/6=sin2x+根号3cos2x=2sin(2x+π/3)
T=π
(2)当x∈[- π/6,π/6]时,2x+π/3∈[ 0,2π/3]所以当x=-π/6有最小值为2sin0=0
答
1)f(x)=sin(2x+π/3)+sin(π/2-2x+π/6)
=sin(2x+π/3)+sin(-2x+2π/3)
=sin(2x+π/3)+sin(2x+π/3)
=2sin(2x+π/3)
T=2π/2=π
2)-π/6=