Question

已知函数f(x)=cosx的4次方+2sinxcosx-sinx的4次方

求f(x)的最小正周期和单调递增区间

Answer 1

f(x)=(cos^2x +sin^2 x)(cos^2 x-sin^2 x)+sin2x
=cos2x+sin2x
=√2sin(2x+π/4)
最小正周期为π
单调增区间为2kπ-π/2=<2x+π/4<=2kπ+π/2, 即kπ-3π/8=<x<=kπ+π/8
这里k为任意整数。

追问

为什么f(x)=(cos^2x +sin^2 x)(cos^2 x-sin^2 x)+sin2x

尤其是后面的sin2x

追答

2sinxcosx=sin2x
(cosx)^4-(sinx)^4=(cos^2x +sin^2 x)(cos^2 x-sin^2 x)

Answer 2

f(x)=2sinxcosx+(cos²x+sin²x)(cos²x-sin²x)
=sin2x+1*cos2x
=√2sin(2x+π/4)
所以T=2π/2=π

递增则2kπ-π/2<2x+π/4<2kπ+π/2
kπ-3π/8<x<kπ+π/8

所以增区间是(kπ-3π/8，kπ+π/8)

Answer 3