高中数学：已知函数f(x)=cos平方x+sinxcosx(x?R),问题在补充里

f[x]=(1+cos2x)/2+1/2·sin2x
=1/2+1/2(cos2x+sin2x)
=1/2+√2/2·sin(2x+π/4)
(1)f(3π/8)=1/2+√2/2·sin(3π/4+π/4)
=1/2.
（2）又2kπ-π/2≤2x+π/4≤2kπ+π/2
解得
kπ-3π/8≤x≤kπ+π/8,
∴f(x)的单调递增区间是[kπ-3π/8,kπ+π/8].

f<x>=cos的平方x加SinxCosx
=(cosx)^2+sinxcosx
=(1+cos2x)/2+1/2sin2x
=1/2[cos2x+sin2x]
=1/2*根号2*sin(2x+pi/4)
所以
f(pi*3/8)
=1/2*根号2*sin(2*3*pi/8+pi/4)
=1/2*根号2*sin(pi)
=0

(2)f(x)=1/2*根号2*sin(2x+pi/4) 的单调递增区间是-pi/2<=2x+pi/4<=pi/2
解得-pi/8<=x<=pi/8
所以单调递增区间为[-pi/8,pi/8]
(pi表示圆周率)