求函数fx=sin(2x-π/4)-2根号2sin²x的单调递减区间及对称轴
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/25 17:53:58
求函数fx=sin(2x-π/4)-2根号2sin²x的单调递减区间及对称轴
求函数fx=sin(2x-π/4)-2根号2sin²x的单调递减区间及对称轴
求函数fx=sin(2x-π/4)-2根号2sin²x的单调递减区间及对称轴
/>
你好
f(x)=sin(2x-π/4)-2√2sin²x
=sin2xcosπ/4-cos2xsinπ/4-√2(1-cos2x)
=√2/2 sin2x- √2/2 cos2x +√2cos2x-√2
=√2/2 sin2x+√2/2 cos2x-√2
=sin(2x+π/4)-√2
函数f(x)=sinx的对称区间是
[2Kπ-...
全部展开
你好
f(x)=sin(2x-π/4)-2√2sin²x
=sin2xcosπ/4-cos2xsinπ/4-√2(1-cos2x)
=√2/2 sin2x- √2/2 cos2x +√2cos2x-√2
=√2/2 sin2x+√2/2 cos2x-√2
=sin(2x+π/4)-√2
函数f(x)=sinx的对称区间是
[2Kπ-π/2,2Kπ+π/2]单调递增
[2Kπ+π/2,2Kπ+3π/2]单调递减
对称轴是Kπ+π/2
所以
2Kπ-π/2≤2x+π/4≤2Kπ+π/2
解得Kπ-3π/8≤x≤Kπ+π/8时,函数单调递增
2Kπ+π/2≤2x+π/4≤2Kπ+3π/2
解得Kπ+π/8≤x≤Kπ+5π/8时,函数单调递减
2x+π/4=Kπ+π/2
对称轴为x=Kπ/2+π/8
很高兴为您解答,祝你学习进步!有不明白的可以追问!
如果有其他问题请另发或点击向我求助,答题不易,请谅解.
如果您认可我的回答,请点击下面的【采纳为满意回答】或者点评价给好评,谢谢!
收起