已知函数f(x)=cosx(sinx-cosx)+1

1、求最小正周期：
f(x)=cosx(sinx-cosx)+1
f(x)=cosx[sinx+sin(3π/2+x)]+1
f(x)=2cosx[sin[(x+3π/2+x)/2]cos[(x-3π/2-x)/2]+1
f(x)=2cosxsin(x+3π/4)cos(-3π/4)+1
f(x)=2cosxsin(x+3π/4)[-(√2)/2]+1
f(x)=1-(√2)cosxsin(x+3π/4)
f(x)=1-(√2)(1/2)[sin(x+3π/4+x)+sin(x+3π/4-x)]
f(x)=1-[(√2)/2][sin(2x+3π/4)+sin(3π/4)]
f(x)=1-[(√2)/2][sin(2x+3π/4)+(√2)/2]
f(x)=1-[(√2)/2]sin(2x+3π/4)-1/2
f(x)=1/2-[(√2)/2]sin(2x+3π/4)
因为：自变量x的系数不是2,
所以：最小正周期是2π/2=π
2、求值域：
因为：-1≤sin(2x+3π/4)≤1
所以：-(√2)/2≤-[(√2)/2]sin(2x+3π/4)≤(√2)/2
因此：(1-√2)/2≤1/2-[(√2)/2]sin(2x+3π/4))≤(1+√2)/2
即：f(x)的值域是：：(1-√2)/2≤f(x)≤(1+√2)/2
3、求单调递减区间：
f(x)=1/2-[(√2)/2]sin(2x+3π/4)
f'(x)=(√2)cos(2x+3π/4)
令：f'(x)＜0,即：(√2)cos(2x+3π/4)＜0
cos(2x+3π/4)＜0
2kπ+π/2＜2x+3π/4＜2kπ+3π/2,其中：k=0、±1、±2、……,下同
2kπ-π/4＜2x＜2kπ+3π/4
kπ-π/8＜x＜kπ+3π/8
即：函数f(x)的单调减区间是：x∈(kπ-π/8,kπ+3π/8),其中：k=0、±1、±2、…….

f(x)=cosx·sinx-cosx·cosx+1=1/2sin2x-1/2cos2x+1/2=1/2sin(2x-四分之一派)+1/2
(1)2 (2)(0,1) (3)单调递减区间是(八分之三派+k派，八分之七派+k派）

f(x)=1/2sin2x-1/2(cos2x+1)+1=√￣2/2sin(2x-π/4)+1/2
（1）T=π (2)【-√￣2/ 2+ 1/2,√￣2/2 + 1/2】
(3)[3/8π,7/8π]

（1）f(x)=cosx(sinx-cosx)+1.
=cosxsinx-cos^2x+1
=1/2sin2x-（cos2x+1）/2+1
=1/2sin2x-1/2cos2x+1/2
=1/2(sin2x-cos2x)+1/2
=√2/2sin(2x-π/4)+1/2
所以：f(x)的最...