题目

求函数f(x) 在区间[-π/4,π/4]的最大值和最小值
f（x）=sin（2x+π/3）+sin（2x-π/3）+2cos²x-1，x∈R

提问时间：2020-11-03

答案

f(x)=sin2xcos(π/3)+cos2xsin(π/3)+sin2xcos(π/3)-cos2xsin(π/3)+cos2x
=2sin2xcos(π/3)+cos2x
=sin2x+cos2x
=√2sin(2x+π/4)
f(max)=√2
f(min)= - √2