Question

已知函数FX=1-COS2X+SIN2X

求函数最小正周期和最大值
求y=fx在R上的单调区间
尽量详细点，谢谢

Answer 1

f(x)=1-cos2x+sin2x
=sin2x-cos2x+1
=√2sin(2x-π/4)+1.。。。。辅助角公式
最小值正周期=2π/2=π
最大值=√2+1

令-π/2+2kπ<=2x-π/4<=π/2+2kπ,k是整数
-π/8+kπ<=x<=3π/8+kπ,k是整数
∴增区间是[-π/8+kπ,3π/8+kπ],k是整数

令π/2+2kπ<=2x-π/4<=3π/2+2kπ,k是整数
3π/8+kπ<=x<=7π/8+kπ,k是整数
∴减区间是[3π/8+kπ,7π/8+kπ],k是整数
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Answer 2

解：
f(x)=1-cos2x+sin2x
=1+sin2x-cos2x
=根号2sin(2x-π/4) + 1
所以最小正周期T=2π/2=π
最大值为根号2 + 1
单调增区间:
-π/2+2kπ<=2x-π/4<=π/2+2kπ
-π/8+kπ<=x<=3π/8+kπ
x∈[-π/8+kπ,3π/8+kπ]
单调减区间:
π/2+2kπ<=2x-π/4<=3π/2+2kπ
3π/8+lπ<=x<=7π/8+kπ
x∈[3π/8+lπ,7π/8+kπ]

Answer 3