函数f(x)=sin(2x+π/6)+cos(2x+π/3)的最小正周期和最大值
问题描述：
函数f(x)=sin(2x+π/6)+cos(2x+π/3)的最小正周期和最大值
答
f(x)=sin(2x+π/6)+cos(2x+π/3)
=√2sina[（2x+π/6)+π/4]
T=π
fmax=√2
答
sin(2x+π/6)+cos(2x+π/3) =√2sin(2x+π/6+π/4) =√2sin(2x+5π/12) （公式AsinX+BcosX=√(A^2+B^2) sin(X+ARGTAN（B/A）?T=2π/2=π,2x+5π/12=π/2+2kπ时有最大值,x=1/24π+kπ,f(x)=√2