阿Q吧 > 函数y=log1⼀2(sin2xcos45-cos2xsin45)单调递减区间

函数y=log1⼀2(sin2xcos45-cos2xsin45)单调递减区间

2025-06-27 01:22:30

推荐回答（2个）

回答1：

解：y=log1/2(sin2xcos45-cos2xsin45)
=log1/2(sin2xcos45-cos2xsin45)
=log1/2[sin(2x-45°)]
又由sin(2x-45°)＞0且y=log1/2(sin2xcos45-cos2xsin45)单调递减区间
当2kπ+π/2≤2x-π/4＜2kπ+π，k属于Z时，y=log1/2(sin2xcos45-cos2xsin45)单调递减
即kπ+3π/8≤x＜kπ+5π/8，k属于Z时，y=log1/2(sin2xcos45-cos2xsin45)单调递减
故函数的减区间为[kπ+3π/8，kπ+5π/8]，k属于Z

回答2：

解y=log1/2(sin2xcos45-cos2xsin45)
=log1/2(sin2xcos45-cos2xsin45)
=log1/2[sin(2x-45°)]
又由sin(2x-45°)＞0且y=log1/2(sin2xcos45-cos2xsin45)单调递减区间
当2kπ+π/2≤2x-π/4＜2kπ+π，k属于Z时，y=log1/2(sin2xcos45-cos2xsin45)单调递减
即kπ+3π/8≤x＜kπ+5π/8，k属于Z时，y=log1/2(sin2xcos45-cos2xsin45)单调递减
故函数的减区间为[kπ+3π/8，kπ+5π/8]，k属于Z