Question

函数y=sin(π4-2x)的单调递减区间为______.

函数y=sin(π4-2x)的单调递减区间为______.

Answer 1

由题意可得:y=sin(
π
4
-2x
)=-sin(2x-
π
4
),
由正弦函数的单调性可知y=sin(2x-
π
4
)的单调增区间为
[2kπ-
π
2
,2kπ+
π
2
],k∈Z
即
[kπ-
π
8
,kπ+
3π
8
],k∈Z
所以y=sin(
π
4
-2x
)=-sin(2x-
π
4
)的减区间为
[kπ-
π
8
,kπ+
3π
8
].k∈Z
故答案为:[-
π
8
+kπ,
3π
8
+kπ](k∈z).