Question

已知函数f（x）=2cosx(sinx-cosx)+1,x属于R。（1）求函数f(x)的最小正周期;(2)求函数f(x)在区间〔π/8，3

已知函数f（x）=2cosx(sinx-cosx)+1,x属于R。（1）求函数f(x)的最小正周期;(2)求函数f(x)在区间〔π/8，3π/4]上的最小值和最大值
过程详细点~~~最好解释下

Answer 1

解：
f(x)=2cosx(sinx-cosx)+1
=2sinxcosx-2(cosx)^2+1
=2sinxcosx-[2(cosx)^2-1]
=sin2x-cos2x
=√2(√2/2*sin2x-√2/2*cos2x)
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
（1）
f(x)=√2sin(2x-π/4)
∴函数f(x)的最小正周期:
T=2π/2=π

（2）f(x)=2sinxcosx+1-2cosx^2
=sin2x-cos2x
=√2sin(2x-π/4)
π/8≤x≤3π/4
得 0≤2x-π/4≤5π/4
f(x）最小值是＝√2sin5π/4=-1
f(x）最大值是＝√2 2sinπ/2=√2

Answer 2

f(x)=2cosx(sinx-cosx)+1
=2sinxcosx-2(cosx)^2+1
=2sinxcosx-[2(cosx)^2-1]
=sin2x-cos2x
=√2(√2/2*sin2x-√2/2*cos2x)
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
（1）
f(x)=√2sin(2x-π/4)
T=2π/2=π

（2）f(x)=2sinxcosx+1-2cosx^2
=sin2x-cos2x
=√2sin(2x-π/4)
π/8≤x≤3π/4
得 0≤2x-π/4≤5π/4
f(x）最小值是＝√2sin5π/4=-1
f(x）最大值是＝√2 2sinπ/2=√2

Answer 3

我现在怀疑你是不是二十二中的，我也在写