函数f(x)=cos(2x-π/3)+cos(2x+π/6)的最大值和最小正周期以及化简过程
问题描述:
函数f(x)=cos(2x-π/3)+cos(2x+π/6)的最大值和最小正周期以及化简过程
答
f(x)=cos(2x-π/3)+cos(2x+π/6)
=cos(2x-π/3)+sin(-2x+π/3)
=cos(2x-π/3)-sin(2x-π/3)
=√ 2cos(2x-π/3+π/4)
=√ 2cos(2x-π/12)
所以函数f(x)的最大值为 √2,最小正周期为π为什么cos(2x+π/6)=sin(-2x+π/3)公式sina=sin(π/2-a)cos(2x+π/6)=sin(π/2-2x-π/6)=sin(-2x+π/3)