求sin2x+cos2x的单调区间

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 04:36:35

求sin2x+cos2x的单调区间
求sin2x+cos2x的单调区间

求sin2x+cos2x的单调区间
因为根据公式asinx+bcosx=根号a^2+b^2sin(x+θ) ∴sin2x+cos2x=根号2sin(2x+π/4) ∴-π/2+2kπ≤2x+π/4≤π/2+2kπ k∈z ∴-3π/8+kπ≤x≤π/8+kπ k∈z π/2+2kπ≤2x+π/4≤3π/2+2kπ k∈z ∴π/8+kπ≤x≤5π/8+kπ k∈z ∴单调增区间为[-3π/8+kπ,π/8+kπ] k∈z 单调减区间为[π/8+kπ,5π/8+kπ] k∈z

=根号2(根号2/2*sin2x + 根号2/2*cos2x)
=根号2sin(2x +π/4 )

π/2+2kπ≤2x+π/4≤3π/2+2kπ
π/4+2kπ≤2x≤5π/4+2kπ
π/8+kπ≤x≤5π/8+kπ
[π/8+kπ,5π/8+kπ]
为单调递减区间
y=sin2x+cos2x=(根号2...

全部展开

=根号2(根号2/2*sin2x + 根号2/2*cos2x)
=根号2sin(2x +π/4 )

π/2+2kπ≤2x+π/4≤3π/2+2kπ
π/4+2kπ≤2x≤5π/4+2kπ
π/8+kπ≤x≤5π/8+kπ
[π/8+kπ,5π/8+kπ]
为单调递减区间
y=sin2x+cos2x=(根号2)sin(2x+π/4)
递增时2kπ-π/2<2x+π/4<2kπ+π/4
kπ-3π/8/所以递增区间是(kπ-3π/8,kπ+π/8)(k∈Z)

收起