Question

已知函数f(x)=2cosx(sinx+cosx).

(1)求f(5兀/4)的值。（2）求函数f(x)的最小周期和单调递增区间。

Answer 1

f(x)=2cosxsinx+2cos方x
=sin2x+1+cos2x
=sin2x+cos2x +1
=√2 sin(2x+π/4)+1

(1)f(5π/4)=sin5π/2+cos5π/2 +1
=1+0+1
=2
(2) T=2π/2=π
增区间满足：
2kπ-π/2≤2x+π/4≤2kπ+π/2
2kπ-3π/4≤2x≤2kπ+π/4
kπ-3π/8≤x≤kπ+π/8
即增区间为：
【kπ-3π/8，kπ+π/8】，k∈Z

Answer 2

f(x)=2cosx(sinx+cosx)
=sin2x+cos2x+1
=√2sin(2x+π/4)+1

f(5π/4)=√2sin(5π/2+π/4)+1=√2sin(3π/4)+1=1+1=2
2)T=2π/2=π
2x+π/4在[2kπ-π/2,2kπ+π/2]上单调递增

x在[kπ-3π/8,kπ+π/8]上单调递增