题目

函数f(x)=2cos^2x+2sinxcosx,（1）求f(x)的最小正周期（2)记函数g(x)=f(x)*f(x+π/4),求函数g(x)的值域.

提问时间：2020-12-21

答案

(1)f(x)=2cos²x+2sinxcosx
=1+cos2x+sin2x
=1+√2(√2/2cos2x+√2/2sin2x)
=1+√2sin(2x+π/4)
f(x)的最小正周期为：2π/2=π
(2) g(x)=f(x)*f(x+π/4)
=[1+√2sin(2x+π/4)][1+√2sin(2x+π/4+π/2)]
=[1+√2sin(2x+π/4)][1-√2cos(2x+π/4)]
=1+√2sin(2x+π/4)-√2cos(2x+π/4)-2sin(2x+π/4)cos(2x+π/4)
=(sin(2x+π/4)-cos(2x+π/4))²+√2(sin(2x+π/4)-cos(2x+π/4))
=(sin(2x+π/4)-cos(2x+π/4)+√2/2)²-1/2
=(√2sin(2x+π/4-π/4)+√2/2)²-1/2
=(√2sin(2x)+√2/2)²-1/2
=2(sin(2x)+1/2)²-1/2
当 -1≤sin2x≤1
g(x)的值域为：-1/2≤g(x)≤4