化简f(x)=2sin(x+ π/2 )sin(x+ π/3)+cos(5π/6),(x属于R）,并求函数f(x)的值域和最小正周期?

化简f(x)=2sin(x+ π/2 )sin(x+ π/3)+cos(5π/6),(x属于R）,并求函数f(x)的值域和最小正周期?
化简f(x)=2sin(x+ π/2 )sin(x+ π/3)+cos(5π/6),(x属于R）,并求函数f(x)的值域和最小正周期?

化简f(x)=2sin(x+ π/2 )sin(x+ π/3)+cos(5π/6),(x属于R）,并求函数f(x)的值域和最小正周期?
∵-1<=sin(2x+π/6)<=1 ∴-1/4<=f(x)<=3/4 即f(x)的值域为：[-1/4,3/4] (2) f'(x)=cos(2x+π/6) f'(x0)=cos(2x0+π

f(x)=2cosx(sinxcosπ/3+cosxsin π/3)+cos(π/2+π/3)=cosxsinx+(3^0.5)(cosx)^2-sinπ/3
=0.5sin2x+(3^0.5)(cosx)^2-sinπ/3=0.5sin2x+0.5*(3^0.5)(1+cos2x)-sinπ/3
=sin(2x+π/3)
最小正周期： π
很久没再看过三角函数了，记忆中的公式解出来的