Question

已知f(x)=(sinx+cosx)²+2cos²x,x∈R,求(1)函数f(x)的最小正周期

求函数的单调减区间，求函数的最小值及取最小值的x集合

Answer 1

f(x)=(sinx+cosx)²+2cos²x
=sin²x+2sinxcosx+cos²x+2cos²x
=1+2sinxcosx+2cos²x
=2sincosx+2cos²x-1+2
=sin2x+cos2x+2........正弦，余弦二倍角公式
=√2sin(2x+π/4)+2.......辅助角公式
令π/2+2kπ<=2x+π/4<=3π/2+2kπ,k∈Z
∴π/8+kπ<=x<=5π/8+kπ,k∈Z
f(x)减区间是[π/8+kπ,5π/8+kπ],k∈Z
f(x)最小值=2-√2
此时
2x+π/4=3π/2+2kπ
x=5π/8+kπ,k∈Z
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