已知函数f(x)=sin(x+π/4)+2cos(x-π/4)+2sin2x+3cos(x+3π/4) ; g(x)=f(x)+f^2(x/2)

题目

已知函数f(x)=sin(x+π/4)+2cos(x-π/4)+2sin2x+3cos(x+3π/4) ; g(x)=f(x)+f^2(x/2)

求f(π/4)

求函数g(x)的最小正周期和单调递增区间

答案

f(x)=√2/2*sinx+√2/2*cosx+2(cosx*√2/2+sinx*√2/2)+2sin2x+3[cosx*(-√2/2)-sinx*√2/2]
=2sin2x
(1) f(π/4)=2sinπ/2=2
(2) g(x)=2sin2x+(2sinx)^2
=2sin2x+4(sinx)^2
=2sin2x+2(1-cos2x)
=2(sin2x-cos2x)+2
=2√2sin(2x-π/4)+2
所以最小正周期T=2π/2=π
令-π/2+2kπ≤2x-π/4≤π/2+2kπ,解得：-π/8+kπ≤x≤3π/8+kπ
所以单调递增区间为：[-π/8+kπ,3π/8+kπ] (k∈Z)