Question

已知函数f(x)=cos(2π-x) cos(π/2-x)-sin^2x (1)求函数f(x)的最小正周期

（2）当x∈[-π/8,3/8π]时，求函数f(x)的值域

Answer 1

解：1、f(x)=cos(2π-x) cos(π/2-x)-sin^2x
=-cosxsinx-sin^2x
=-½sin2x-(1-cos2x)/2
=-1/2sin2x+1/2cos2x-1/2
= - √2/2sin(2x-π/4)-1/2
所以函数f(x)的最小正周期为π
2、当x∈[-π/8,3π/8]时，-π/2≤2x-π/4≤π/2 ,f(x）最小值为x=3π/8时，f(x)=-1/2-√2/2
x=π/8时f(x)最大值为√2/2-1/2
所以当x∈[-π/8,3/8π]时，函数f(x)的值域为[-1/2-√2/2，√2/2-1/2]

Answer 2

1、f(x)=cos(2π-x) cos(π/2-x)-sin^2x
=cosx*sinx+1/2(1-2sin^2x )-1/2
=1/2sin2x+1/2cos2x-1/2
=√2/2(sin2xcosπ/4+cos2xsinπ/4)-1/2
=√2/2sin(2x+π/4)-1/2
所以函数f(x)的最小正周期为π
2、当x∈[-π/8,3/8π]时，f(x）最小值为x=-π/8或3/8π时，f(x)=-1/2
x=π/8时f(x)最大值为√2/2-1/2
所以当x∈[-π/8,3/8π]时，函数f(x)的值域为[-1/2，√2/2-1/2]

Answer 3

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